n--n_«_n_n_n_n_n_n^ 


REESE  LIBRARY 


UNIVERSITY  OF  CALIFORNIA. 


Class 


Bridge  and  Structural  Design 


BY 


W.    CHASE    THOMSON, 

M.  Can.  Soc.  C.  E. 

Assistant  Engineer  Dominion  Bridge  Co., 
Montreal,  Canada. 


NEW  YORK  : 

THE  ENGINEERING  NEWS  PUBLISHING  COMPANY. 
1905. 


Copyright,  1905,  by 
The  Engineering  News  Publishing  Company. 


PREFACE. 

This  book  has  been  developed  from  lectures  given  by  the  author 
during-  the  past  five  years,  under  the  auspices  of  the  Dominion 
Bridge  Co.  His  object  has  been  to  teach  the  elements  of  bridge 
and  structural  design  in  a  simple  and  practical  manner.  Arts.  I  to 
14  inclusive  treat  of  the  general  principles  of  design,  and  are  illu- 
strated by  numerous  examples  ;  while  the  remaining  Articles  are 
examples  of  typical  structures,  in  which  the  stresses  are  analyzed, 
the  members  proportioned,  and  the  details  carefully  worked  out. 

Both  analytical  and  graphical  methods  have  been  employed 
for  obtaining  stresses,  and  the  one  which  seemed  best  suited  for 
any  particular  subject  has  been  adopted.  But  few  tables  are  given, 
as  it  was  thought  unnecessary  to  repeat  information  given  in  any 
of  the  rolling  mills'  hand-books. 

Although  the  book  is  intended  principally  for  students  and 
draughtsmen,  there  are  parts  which  may  be  of  interest  to  practicing 
bridge  designers.  Particular  attention  is  here  drawn  to  Art.  17, 
which  treats  of  the  design  of  a  knee-braced  mill  building ;  and  to 
Art.  19  which  discusses  the  rivet  spacing  and  web  splices  in 
plate  girders,  in  which  one-eighth  of  the  web  plate  is  counted  on 
as  flange  area. 

MONTREAL,  March  10,  1905. 


CONTENTS. 

Page 

Art.    I.  Definitions i 

2.  The  Composition  and  Resolution  of  Forces 2-3 

3.  Examples  in  Graphical  Statics 4-6 

4.  The  Lever  and  Moments 7-8 

5.  Shearing  and  Bending-  Stresses  in  Beams 9-10 

6.  Moment  of  Resistance 1 1 

7.  Moment  of  Inertia 12 

8.  Radius  of  Gyration 13 

9.  Formulae  Relating  to  Beams 14 

10.  Examples  in  the  Computation  of  Properties  of  Simple 

and  Compound  Sections 14-19 

11.  Examples    Illustrating  the   Method  of   Determining 

the  Sizes  of  Beams 20 

12.  Columns  and  Struts 21-23 

13.  Examples  Illustrating  Method  of  Designing  Columns 

and  Struts 24 

14.  Rivets  and  Riveting 25-26 

15.  The  Complete  Design  of  a  Roof  Truss  for  Building 

with  Masonry  or  Brick  Walls  Capable  of  With- 
standing Wind  Pressure 27-32 

16.  Roof  Trusses  Supported  by  Steel  Columns 33 

17.  The  Design  of  a  Knee-Braced  Mill  Building 34-41 

18.  The  Design  of  a  Plate  Girder 42-49 

19.  Plate  Girder  with  One-Eighth  of  Web  Plate  Com- 

puted as  Flange  Area 5O~53 

20.  Design  for  a  Warren  Girder  Highway  Bridge 54~6i 

21.  Design  for  Skew  Warren  Girder  Highway  Bridge. .  .62-67 

22.  Design   for  a  Pin-Connected  Pratt  Truss  Highway 

Span 68-88 


BRIDGE  AND    STRUCTURAL    DESIGN, 

Bv  W.  CHASE  THOMSON,  M.  Can,  Soc.  C.  E. 


ART.  1.— DEFINITIONS. 

Mechanics  is  the  study  of  the  effect  of  force  upon  matter. 

Force  is  the  action  of  gravity,  wind,  steam,  etc.,  causing  or  tend- 
ing to  cause  motion. 

Matter  is  any  substance  whatever,  as  metal,  stone,  wood,  water, 
air,  etc. 

A  body  is  any  piece  of  matter,  and  its  weight  is  the  amount  of 
force  which  gravity  exerts  upon  it. 

Statics  is  that  branch  of  mechanics  which  investigates  the  stresses 
in  a  body  produced  by  forces  which  keep  it  stationary,  as  in  bridges 
and  buildings. 

Dynamics,  on  the  other  hand,  investigates  forces  which  move  the 
body  upon  which  they  act,  as  in  engines  and  other  machinery. 

Stress  is  the  effect  produced  by  equal  and  opposite  forces,  and  is 
measured  in  pounds  or  tons.  It  is  equal  to  only  one  of  the  forces, 
however.  Thus,  if  two  men  pull  40  Ibs.  each  at  opposite  ends  of  a 
rope,  the  stress  will  not  be  80  Ibs.,  but  only  40  Ibs. ;  and  if  a  load  of 
1,000  Ibs.  rest  on  top  of  a  column,  then  the  reaction  of  the  founda- 
tion on  which  the  column  rests  will  also  be  1,000  Ibs.,  but  the  stress 
in  column  will  be  equal  to  but  one  of  these  forces,  viz.:  1,000  Ibs. 
There  can  be  no  stress  without  a  reaction  which  is  always  equal  to 
the  force  acting  on  body. 

Tensile  Stress  or  Tension  occurs  when  two  opposing  forces  tend 
to  pull  apart  the  body  upon  which  they  act,  as  in  the  tie  bars  and 
bottom  chords  of  a  bridge. 

Compressive  Stress  or  Compression  occurs  when  the  opposing 
forces  acting  on  a  body  tend  to  compress  it,  as  in  the  posts  and 
top  chords  of  a  bridge. 


2  BRIDGE  AND  STRUCTURAL  DESIGN. 

ART.  2.— THE  COMPOSITION  AND  RESOLUTION  OF  FORCES. 

The  resultant  of  two  or  more  forces  is  a  single  force  which  will 
produce  the  same  effect  as  the  combined  action  of  those  forces. 

The  components  of  a  force  are  the  several  forces  which  by  their 
combined  action  would  have  the  same  effect  as  the  single  force. 

Let  A  B  and  C  B  in  Fig.  i  represent  both  the  magnitude  and 
direction  of  two  forces  in  the  same  plane,  acting  through  the  point 
B.  Through  A  and  C  lines  are  drawn  parallel  to  the  forces,  inter- 
secting in  the  point  D.  Then  the  diagonal  B  D  represents  both 
the  magnitude  and  direction  of  the  resultant  of  the  forces  A  B  and 
C  B.  A  force  of  equal  magnitude  but  acting  in  the  opposite  direc- 
tion would  balance  the  original  forces,  or,  in  other  words,  the  three 
forces  would  be  in  equilibrium.  The  figure  is  called  the  parallelo- 
gram of  forces. 


Fig.  Z 


In  Fig.  2  it  is  required  to  find  the  components  of  the  force  repre- 
sented by  the  load  W  applied  at  the  apex  of  the  two  rafters  A  B 
and  B  C.  A  vertical  line  B  D  is  drawn  equal  to  the  load  W,  and 
from  the  point  D  lines  are  drawn  parallel  to  the  rafters.  Then  D 
E,  parallel  to  the  rafter  A  B,  is  the  stress  in  that  member,  and  F  D 
the  stress  in  B  C.  The  horizontal  lines  F  H  and  G  E  represent  the 
horizontal  thrust  of  rafters  which  is  the  same  for  both.  B  H  is  the 
vertical  thrust  of  rafter  A  B,  and  B  G  the  vertical  thrust  of  rafter 
B  C. 

In  Fig.  3  one  rafter  A  B  is  inclined,  and  the  other  rafter  B  C  is 
horizontal.  The  force  W  is  applied  at  B.  B  D  is  drawn  vertically 
as  before  equal  to  the  load  W,  and  from  the  point  D  lines  parallel  to 
A  B  and  B  C  are  drawn.  F  D  the  horizontal  component  of  the 
stress  in  A  B  is  equal  to  the  stress  in  B  C.  B  D  the  vertical  com- 
ponent of  the  stress  in  A  B  is  equal  to  the  load.  In  other  words, 
the  rafter  A  B  takes  the  whole  vertical  load,  and  B  C  only  horizontal 
thrust. 

If  any  number  of  forces  in  the  same  plane  meet  in  a  point  as  in 


THE  COMPOSITION  AND  RESOLUTION  OF. FORCES. 


Fig.  4,  their  resultant  may  be  found  by  drawing  lines  end  to  end 
equal  and  parallel  to  the  forces  as  in  Fig.  4a.  The  closing  line  rep- 
resents the  magnitude  and  direction  of  the  resultant.  This  diagram 
is  called  the  polygon  of  forces.  The  arrow  heads  represent  the 
direction  of  the  forces  and  follow  each  other  around  the  diagram. 
The  resultant,  however,  always  acts  towards  the  last  force  drawn. 
A  force  equal  to  the  resultant  but  acting  in  the  opposite  direction 
would  hold  the  other  forces  in  equilibrium. 

If  any  number  of  forces  in  the  same  plane  and  acting  through  the 
same  point  are  in  equilibrium,  their  force  polygon  will  form  a  closed 
figure,  and  if  the  direction  of  all  the  forces  be  known  and  the  mag- 
nitude of  all  but  two,  these  may  be  found  by  scale  from  the  force 


C  75- 
^ 
B 


fig.  5 


polygon.  This  case  is  similar  to  the  forces  acting  at  the  panel 
points  of  a  bridge  or  roof  truss,  and  the  principle  enables  one  to 
find  the  stresses  in  all  the  members  of  a  framed  structure  by  the 
graphical  method. 

In  Fig.  5  there  are  five  forces  acting  through  the  same  point 
which  are  denoted  by  the  letters  between  which  they  lie.  This  is 
the  usual  and  undoubtedly  the  best  method  of  denoting  stresses 
when  solving  them  by  the  graphical  method.  The  stresses  A  B  = 
5,000  Ibs.,  B  C  =  12,000  Ibs.  and  C  D  25,000  Ibs.  are  given,  and  the 
direction  only  of  D  E  and  E  A. 

Fig.  5a  is  the  force  polygon  which  is  constructed  by  drawing 
lines  parallel  and  equal  to  the  forces  A  B,  B  C  and  C  D.  From  the 
point  d  in  force  polygon  a  line  is  drawn  parallel  to  the  force  D  E 
but  of  indefinite  length ;  another  line  is  drawn  from  the  point  a  par- 
allel to  E  A.  The  intersection  of  the  two  lines  in  the  point  e  deter- 


4  BRIDGE  AND  STRUCTURAL  DESIGN, 

mines  the  magnitude  of  the  forces  D  E  and  E  A.  D  E  is  about 
9,000  Ibs.  and  E  A  about  35,000  Ibs.  Any  force  acting  away  from 
the  point  indicates  tension,  and  any  force  acting  towards  it,  com- 
pression. 

ART.    3.— EXAMPLES    IN    GRAPHICAL    STATICS. 

Fig.  6  represents  a  simple  roof  truss.  The  span  is  20  ft.  depth  at 
centre  5  ft.,  and  trusses  spaced  10  ft.,  centre  to  centre.  Assumed 
load  50  Ibs.  per  square  foot  of  horizontal  projection,  concentrated 
at  panel  points  by  purlins.  The  total  load  on  truss  will  be  20'  x  10' 
x  50  Ibs.  =  10,000  Ibs.  The  load  at  each  intermediate  panel  point 
10,000 

= =  2,500  Ibs.,  and  at  each  end  panel  point,  one-half  of 

4 

this  amount  =  1,250  Ibs.  The  end  panel  loads  are  carried  directly 
by  the  walls,  and  therefore  have  no  effect  on  the  stresses  in  the 
truss.  Capital  letters  are  used  to  denote  the  external  forces  which 
consist  of  loads  and  the  reactions  of  the  end  supports ;  and  small 
letters  for  the  internal  forces.  The  truss  members  are  indicated  by 
the  letters  between  which  they  lie. 

Fig.  6a  is  the  stress  diagram.  Beginning  at  the  left  hand  end  of 
truss  and  going  around  it  in  a  right  handed  direction,  as  indicated 
by  curved  arrow,  the  forces  A  B,  B  C,  C  D,  D  E  and  E  F  are  laid 
off  downwards  on  the  vertical  load  line,  Fig.  6a.  The  next  external 
force  is  the  reaction  of  the  right-hand  support  equal  to  one-half 
the  load  on  span.  This  force  is  laid  off  upwards  from  F  to  G  as  it 
acts  in  the  opposite  direction  to  the  loads.  Finally,  the  reaction  of 
the  left-hand  support  is  laid  off  from  G  to  A  the  point  of  beginning. 
Now,  at  the  left-hand  end  of  truss  there  are  four  forces  meeting  in  a 
point.  Two  of  these  forces,  the  reaction  G  A  and  the  load  A  B  are 
known ;  and  two  forces,  the  stresses  in  rafter  Bb  and  bottom  chord 
bG,  are  unknown.  From  the  point  B  on  load  line,  a  line  is  drawn 
parallel  with  the  rafter  Bb,  and  from  G,  a  line  parallel  with  the  bot- 
tom chord  bG.  These  two  lines  intersect  in  the  point  b,  and  deter- 
mine the  stresses  Bb  and  bG.  Next,  at  the  first  panel  point  from 
the  end,  the  stress  in  bB  has  just  been  found,  the  load  BC  is  known 
and  the  stresses  in  Cc  and  cb  unknown.  From  the  point  C  on  load 
line,  a  line  is  drawn  parallel  to  the  rafter  Cc,  and  from  the  point  b  in 
stress  diagram,  a  line  parallel  to  the  strut  cb.  The  two  intersect  in 
the  point  e,  and  determine  the  stresses  in  Cc  and  cb.  At  the  apex 


EXAMPLES  IN  GRAPHICAL   STA  TTCS. 


of  rafters  there  are  now  two  unknown  forces,  the  stresses  in  Dd 
and  dc.  From  the  point  D  on  load  line,  a  line  is  drawn  parallel  to 
Dd,  and  from  the  point  c  in  stress  diagram,  a  line  parallel  to  dc. 
The  two  intersect  in  the  point  d  and  determine  the  stresses  in  Dd 
and  dc. 

It  is  unnecessary  -to  proceed  any  further  with  the  stress  diagram, 
as  the  stresses  in  the  right-hand  end  of  truss  will  evidently  be  the 
same  as  those  in  the  left-hand  end,  but  to  test  the  accuracy  of  the 
work  it  is  sometimes  advisable  to  go  through  the  whole  truss,  and 
if  the  work  has  been  carefully  done  the  stress  diagram  will  form  a 
closed  figure. 

Now  to  know  whether  a  member  is  in  compression  or  tension,  it 


is  necessary  to  observe  which  way  the  stresses  act  in  the  stress  dia- 
gram. In  going  around  any  panel  point  in  the  direction  in  which 
the  external  forces  have  been  taken  (in  this  case  from  left  to  right) 
if  a  force  in  the  stress  diagram  acts  towards  the  panel  point,  the 
member  is  in  compression,  and  if  away  from  it,  in  tension.  For  ex- 
ample :  At  the  apex  of  rafters,  going  around  the  point  from  left  to 
right,  and  observing  the  direction  of  the  forces  in  the  stress  dia- 
gram, it  will  be  seen  that  cC  acts  towards  the  point,  and  therefore 
the  stress  in  rafter  cC  is  compression ;  the  load  C  D  acts  towards 
the  point,  and  the  force  Dd  acts  towards  it.  The  next  force  dc  acts 
away  from  the  point,  and  so  the  stress  dc  is  tension. 

The  external  forces  and  stresses  are  shown  on  the  truss  diagram, 


6  BRIDGE  AND  STRUCTURAL  DESIGN. 

Fig.  6.  The  sign  (-f)  indicates  compression,  and  the  sign  ( — ) 
tension. 

One  of  the  commonest  forms  of  roof  trusses  is  that  known  as  the 
Fink  Truss,  so-called  from  the  name  of  the  inventor.  Fig.  7  is  an 
example. 

The  span  is  40  ft.,  the  angle  of  roof  with  horizontal  30°,  the  panel 
load  2,500  Ibs.  The  half-panel  loads  at  the  ends  have  been  omitted, 
as  they  have  no  effect  on  the  stresses.  Beginning  at  the  left-hand 
end  of  truss,  as  in  previous  example,  the  loads  A  to  H  are  laid  off 
on  the  load  line  in  Fig.  7a  downwards,  and  the  reactions  H  I  and 
I  A  upwards  to  the  point  of  beginning.  The  stress  diagram  is  then 


proceeded  with,  beginning  at  left  end.  At  panel  point  B  C  a  slight 
difficulty  is  encountered,  where  there  are  three  unknown  forces,  viz., 
Cc,  cc,  and  ct  b,  and  the  diagram  cannot  be  completed  when  there 
are  more  than  two  unknown.  At  the  lower  end  of  strut  b,  c,  the 
same  difficulty  is  met  with.  A  very  nice  method  of  solving  this 
problem  is  to  change  some  of  the  web  members  temporarily  as  in 
Fig.  8,  from  which  the  stress  diagram,  Fig.  8a,  is  obtained,  and  the 
stress  in  the  bottom  chord  member  i  I.  The  web  members  may 
then  be  changed  back  to  their  original  form,  and  the  polygon  of 
forces  completed  for  the  panel  point  at  lower  end  of  strut  b,  d 
where  there  are  now  only  two  unknown  forces,  viz.,  bl  C,  and  c,  i. 


THE  LEVER  AND  MOMENTS.  7 

After  which  the  polygon  of  forces  for  panel  point  B  C  may  be 
drawn.  The  rest  is  simple. 

When  the  truss  is  symmetrical  and  the  panel  loads  equal,  as  in 
the  present  example,  there  is  no  difficulty  in  constructing  the  stress 
diagram,  as  the  points  a,  b,  c,  d  will  always  lie  in  a  straight  line; 
but  if  the  panel  lengths  or  the  loads  are  unequal,  as  is  sometimes 
the  case,  it  will  be  necessary  to  use  some  method  for  finding  an 
extra  force  either  at  the  upper  or  lower  end  of  strut  b,  ct. 

One-half  of  the  stress  diagram  only  has  been  constructed,  as  the 
other  half  would  be  exactly  the  same. 

A  common  form  of  roof  truss  is  shown  in  Fig.  9.  The  span  is  50 
ft.,  centre  to  centre,  the  depth  at  ends  4  ft.  and  at  centre  6  ft.  The 
intermediate  panel  loads  are  2,500  Ibs.  each,  and  the  end  panel  loads 
1,250  Ibs.  each. 

Fig.  Qa  is  the  stress  diagram  which  is  only  constructed  for  one- 
half  the  truss.  There  is  no  stress  in  the  end  panels  of  bottom  chord 
oM  and  5M  from  the  vertical  loads.  These  members  are  required 
for  lateral  stability. 

Sometimes  a  roof  truss  is  required  to  slope  in  one  direction  only 
as  in  Fig.  10.  The  stresses  are  found  in  the  same  manner  as  before, 
but  it  is  necessary  to  make  the  stress  diagram  for  the  whole  truss. 

The  span  is  40  ft.  the  depth  at  one  end  4  ft.  and  at  the  other  end 
8  ft.  The  intermediate  panel  loads  are  2,500  Ibs.  each,  and  the  end 
panel  loads  1,250  Ibs.  each. 

ART.  4.— THE  LEVER  AND  MOMENTS. 

If  a  force  act  on  a  body  tending  to  rotate  it  about  a  certain  point, 
it  is  said  to  have  a  moment  about  that  point  equal  to  the  amount  of 
the  force  multiplied  by  the  perpendicular  distance  from  the  line  of 
action  of  force  to  the  said  point. 

In  Fig.  ii  the  force  F  acts  about  the  point  a  with  a  leverage 
equal  to  ab.  The  point  a  is  called  the  point  of  moments,  and  the 
distance  ab  the  lever  arm. 

There  may  be  two  or  more  forces  tending  to  rotate  a  body  about 
the  same  point  either  in  the  same  or  in  the  opposite  direction ;  and 
if  the  body  is  in  equilibrium  the  sum  of  the  left-hand  moments  must 
be  equal  to  the  sum  of  the  right-hand  moments.  Fig.  12  represents 
a  beam  supported  at  the  point  B.  The  load  at  A  tends  to  rotate 
the  beam  in  a  left-handed  direction  about  its  point  of  support,  and 


8 


BRIDGE  AND  STRUCTURAL  DESIGN. 


its  moment  =  3  Ibs.  x  10'  =  30  ft.-lbs.  The  load  at  C  tends  to  ro- 
tate the  beam  in  a  right-handed  direction,  and  its  moment  =  5  Ibs. 
x  6'  =  30  ft.-lbs.  The  moments,  therefore,  are  equal  but  opposite, 
so  the  beams  will  remain  horizontal.  A  lever  may  either  be  straight 
or  bent,  but  no  matter  what  the  actual  length  of  the  lever  may  be 
the  true  lever  arm  is  the  perpendicular  distance  from  the  line  of 
force  to  the  point  of  moments.  Figs.  13  and  14  are  examples  of 
bent  levers. 

Fig.  15  represents  a  bracket  on  the  side  of  a  wall  supporting  a 
load  W  at  the  point  B.    The  principle  of  the  lever  is  here  employed 


Fig.M 


\vtt*  arm 


* 
levgf  arm        Mgvetar 


Pg^^/x^wt^/X,^^^^^ 
— '    : 


FifctG 


fig.  17 


to  determine  the  stresses.  For  the  stress  in  A  B  moments  are  taken 
about  the  point  C,  then  W  X  lever  arm  C  E  =  stress  in  A  B  X  lever 


arm  A  C.    Therefore  stress  in  A  B  = 


W  xC  E 
A  C 


For  the  stress 


in  C  B  moments  are  taken  about  the  point  A.     Then  WX  lever  arm 
A  B  =  stress  in  C  BX  lever  arm  A  D.     Therefore  stress  in  C  B  = 
W  x  A  B 


A  D 

To  determine  the  reactions  for  a  beam  supported  at  both  ends  and 


SHEARING  AND  BENDING  STRESSES  IN  BEAMS.  g 

loaded  in  any  manner,  moments  of  all  the  loads  are  taken  about  one 
support,  and  divided  by  the  distance  centre  to  centre  of  supports. 

Example :  On  the  beam  A  B,  Fig.  16,  there  are  four  loads  placed 
as  shown.  For  the  reaction  at  A  moments  are  taken  about  B  and 
divided  by  the  span  as  follows  : 

5,ooox   2  =  10,000 

2,000  x    7  =  14,000 

4,000  x  10  =  40,000 

3,500  x  14=  49,000 

113,000  ft.-lbs.  -f-  17'  =  6,647  Iks. 

The  loads  tend  to  rotate  the  beam  about  the  point  B  in  a  left- 
hand  direction  with  a  moment  of  113,000  ft.-lbs.  The  reaction  at  A 
tends  to  rotate  the  beam  about  the  point  B  in  a  right-handed  direc- 
tion with  a  moment  —  6,647  Ibs.  x  17'  =  1 13,000  ft.-lbs.  These  mo- 
ments are  equal  but  opposite,  and  therefore  counteract  each  other, 
so  there  is  no  resultant  moment  at  the  support  B. 

To  obtain  the  reaction  B,  moments  may  be  taken  about  A,  but 
as  the  sum  of  the  reactions  must  be  equal  to  the  sum  of  the  loads,  it 
is  only  necessary  to  add  together  the  loads,  and  subtract  the  reac- 
tion A.  The  result  will  be  the  reaction  B,  thus : 

5,000  +  2,000  +  4,000  +  3,500  —  6,647  =  7>853  —  reaction  B. 


ART.    5.— SHEARING    AND    BENDING    STRESSES    IN    BEAMS. 

A  loaded  beam  is  subjected  to  two  kinds  of  stress,  viz. :  Shearing 
and  bending.  The  shearing  stress  tends  to  cause  the  particles  of 
the  beam  to  slide  by  one  another  in  a  vertical  plane,  as  when  a  plate 
is  cut  in  a  shearing  machine.  Fig.  17  represents  a  beam  loaded 
uniformly  with  a  load  =  w  per  lineal  foot.  At  each  end  there  are 
equal  and  opposite  forces  acting  on  the  beam,  viz.,  one-half  of  the 
load  acting  downwards,  -and  the  reaction  of  the  support  acting  up- 

w  1 
wards,  each  equal  to  .    These  two  forces  tend  to  shear  the 

2 

beam,  or  cut  it  crosswise.    The  shearing  force  at  any  point,  distant 
x  from  one  end,  is  equal  to  the  reaction  at  that  end,  less  the  load  on 

w  1 

the  length  x ;  or,  shear  at  x  = w  x.    The  bending  stresses 

2 
cause  compression  on  the  upper  side  of  the  beam  and  tension  on  the 


10 


BRIDGE  AND  STRUCTURAL  DESIGN. 


B«am   suppotted  and  fi««d  d\  Ooe  end,  and 
a  load  VI  at  olher  end. 


mom«nt  Diagram. 

Max.mom«nt  isal  fixed  «nd,  -VYl 

Momcrf  al  any  point  cTtatanl  x  from  toad  •Htx 


Fig.  is^  .  Shear  D*iagram.  Shear  at  any  poirtf  •  VI. 


Fig.  19.  Beam  supported  and  fixed  al  one  end. and 
carrying  a  toad  to-  per  lineal  foot. 

Fig.idS.  Moment  Diagram.  .,. 

Ma*.  Moment  »s  al  f  wed  end.  •  T*"1- 
Moment  at  any  point  distant  jt  from  ffw  ?nd  «^5 
The  curve  is  a  parabola  with  vertex  a|  free  end. 

Fid.  t9fc.  Shear  Diagram. 

Max.  Shear  is  al  fixed  end.  «  u>£. 

Shear  at  any  point  distant  x  from  free  end.ifx 


—  X    -» 


_^_ 


Fi£.2o.    Beam  supported  at  bof*  end, an*  carnlna  a 
concentrated  load  W  at  centre. 

Fig  2ol.  Moment  Diagram.  . 

Max.  Moment  is  at  centre.*  *$?: 
Moment  at  any  point  x  between  end  and  cenfte.  ^ 


Fi$.2£*  Shear  Dfeg< 


*. 


Fig.  21.    Beam  suppotted  at  both  ends  and  carrying  a 
uniform  load  UP  per  lineal  foot. 

FiA2ia.  Moment  Diagram.  ^ 

Max.  Moment  \s  at  centre.  •  -^r~. 
Moment  at  any  point  distant  x'  from  support. 

The  curve  is  a  parabola  *ith  vertex  at  cento. 

Fig.2i&.  Shear  Diagram. 

Max,  shear  at  ends . »  ^*.  .. 

Shear  at  any  point  distant  %  from  support,  ^-w 
Shear  at  centre  .o. 


MOMENT  OF  RESISTANCE.  1 1 

lower  side.  In  the  case  of  an  open  girder  the  bending  moment  is 
all  resisted  by  the  flanges ;  but,  in  a  solid  beam,  it  is  resisted  by  the 
entire  section. 

Bending  moments  and  shears  for  various  cases  are  illustrated  in 
Figs.  18,  19,  20  and  21. 

ART.    6.— MOMENT    OF    RESISTANCE. 

When  a  beam  is  loaded  transversely,  the  fibres  on  one  side  of  the 
neutral  axis  are  compressed  and  those  on  the  other  side  extended, 
while  the  fibres  in  the  neutral  axis  are  neither  compressed  nor  ex- 
tended, and  the  beam  will  assume  a  curved  form  as  in  Fig.  22.  The 
extreme  outer  fibres  are  stressed  most,  and  the  intermediate  ones 
in  direct  proportion  to  their  distance  from  the  axis. 


Fig.  22 


The  moment  of  resistance  of  a  »/eam  is  the  moment  of  all  the 
fibre  stresses  about  the  neutral  axis. 

The  following  is  a  general  method  for  determining  the  moment 
of  resistance  of  a  beam  of  any  section : 

In  Fig.  23.    f  =  stress  per  square  inch  on  outer  fibres. 

n  =  distance  in  inches  from  neutral  axis  to  outer  fibres. 
y  =  distance  in  inches  from  neutral  axis  to  any  fibre. 
A  =a  small  or  elementary  area. 

Then  —  =  stress  per  square  inch  on  fibres  at  distance  of  one 
n 

inch  from  neutral  axis. 

— y= stress  per  square  inch  on  fibres  at  distance  of  y 
n 

from  neutral  axis. 

— y  A  =  stress  on  an  element  of  fibres  at  distance  of  y 
n 

from  neutral  axis. 

f — yA  )y  =  moment  of  stress  on  an  element  of  fibres 
x  n 

at  distance  of  y  from  neutral  axis. 


12  BRIDGE  AND  STRUCTURAL  DESIGN. 

This  last  expression  taken  for  all  values  of  y  both  above  and 
below  the  neutral  axis  is  the  moment  of  stress  (or  moment  of  re- 
sistance) of  the  given  section,  or 


n  n 

The  factor  (•2>y2A)  is  called  the  moment  of  inertia  and  is  repre- 
sented by  I.  It  is  obtained  by  multiplying:  each  elementary  area 
by  the  square  of  its  distance  from  the  neutral  axis  and  taking:  the 
sum  of  the  products. 

Representing:  (^y2^)  by  I,  then  M=  —  I  =  moment  of  resistance. 

n 

The  factor  (  —  )  of  the  moment  of  resistance  is  usually  represented 

v  n  ' 

by  R  and  called  the  moment  of  resistance,  which  is  not  strictly 
correct,  for  the  real  moment  of  resistance  =Rf. 

I 

—  is  sometimes  called  the  section  modulus  and  represented  by  S. 
n 

The  outer  fibres  only  of  a  beam  receive  the  maximum  stress  per 
square  inch,  hence  the  more  metal  concentrated  in  the  flange,  the 
greater  its  resistance  to  bending. 

ART.  7.—  MOMENT  OF  INERTIA. 

As  stated  in  Art.  6,  the  moment  of  inertia  is  a  factor  of  the  mo- 
ment of  resistance,  and  is  represented  by  I.  The  following  is  an  ap- 
proximate method  for  finding  the  moment  of  inertia  of  a  beam  of 
any  section  about  an  axis  through  its  centre  of  gravity. 

The  section  should  be  divided  into  a  number  of  narrow  strips 
parallel  with  the  neutral  axis,  the  area  of  each  strip  calculated,  and 
the  distance  of  its  centre  line  from  the  neutral  axis  measured.  Each 
area  should  be  multiplied  by  the  square  -of  its  distance  from  the 
neutral  axis,  then  the  sum  of  these  products  will  be  the  approxi- 
mate moment  of  inertia.  The  narrower  the  strips,  the  more  ac- 
curate will  be  the  result,  but  it  will  always  be  a  trifle  too  small.  To 
be  exact,  the  moment  of  inertia  of  each  strip  about  an  axis  through 
its  own  centre  of  gravity  should  be  added  to  the  last  result  —  but 
this  is  unnecessary  in  practice. 

For  the  moment  of  inertia  of  a  rectangle  about  an  axis  through 
its  centre  of  gravity,  the  section  is  supposed  to  be  divided  into  nar- 


RADIUS  OF  G  YRA  TION. 


..  5  J 


row  strips  as  shown  in  Fig.  24.    The  length  of  each  strip  is  b,  its 
thickness=A>  and  the  distance  of  its  centre  from  the  neutral  axis 

=  y- 

Then  the  summation  of  (bAy2)=the  approximate  moment  of 
inertia. 

By  the  help  of  the  calculus,  the  thickness  of  each  strip  can  be 
made  infinitely  small  and  therefore  an  exact  result  obtained  which 

bh3 

is  I  --  .     In  which  b  =  the  width  of  beam  and  h  =  the  height. 
12 

This  is  important. 

The  moment  of  inertia  of  a  triangle  about  an  axis  through  its 
centre  of  gravity  and  parallel  with  the  base,  as  shown  in  Fig.  25,  is 


The  moment  of  inertia  of  a  circle  about  an  axis  through  its  centre 

Tfd4 

of  gravity,  as  shown  in  Fig.  26  is  I  —    —  . 

64 
The  moment  of  inertia  of  a  compound  section  about  an  axis 

through  its  centre  of  gravity  is  equal  to  the  sum  of  the  moments  of 
inertia  of  the  component  parts  about  axes  through  their  own  centres 
of  gravity,  plus  the  areas  of  the  component  parts  multiplied  by  the 
squares  of  the  distances  of  their  centrestof  gravity  from  the  neutral 
axis  of  the  whole  figure. 

ART.    8.—  RADIUS    OF    GYRATION. 

The  radius  of  gyration,  which  is  usually  represented  by  r,  is  the 
distance  from  the  neutral  axis  through  the  centre  of  gravity  of  a 
section  to  a  point  where,  if  the  toial  area  could  be  concentrated  and 
multiplied  by  the  square  of  this  distance,  the  result  would  be  the 
moment  of  inertia  of  the  section  about  the  same  axis  ;  thus  I  =  area 

X  r2,  and  therefore  r=  \  -  •       The  radius  of  gyration  is  used 

area 

principally  in  formulas  for  the  strength  of  columns. 


14  BRIDGE  AND  STRUCTURAL  DESIGN. 

ART.  9.— FORMULAE  RELATING  TO  BEAMS. 

The  following  notation  and  relations  between  bending  moments 
and  the  various  properties  of  beams,  which  have  already  been 
treated  of  in  Arts.  6,  7  and  8,  are  here  set  forth  more  concisely. 
The  student  should  familiarize  himself  with  the  formulae  as  they 
will  be  referred  to  frequently  in  the  following  pages : 

M  =  bending  moment  in  inch-pounds. 

R  =  moment  of  resistance,  =  S.  the  section  modulus. 

f    =  stress  per  square  inch  on  outer  fibres. 

I   —  moment  of  inertia  about  axis  through  centre  of  gravity  of 
section. 

A   =  area  of  section. 

n   =  distance  from  centre  of  gravity  of  section  to  extreme  outer 

fibres, 
r   =  radius  of  gyration. 

Then  M  =  Rf. 


ART.  10.— EXAMPLES  IN  THE  COMPUTATION  OF  PROPERTIES    OF  SIMPLE 
AND    COMPOUND    SECTIONS. 

Fig.  2,7  represents  a  6  X  3^  X  ^  angle.  It  is  divided  into  two 
rectangles:  one  6"  X  .5"  =  30",  and  one  3"  X  .5"  =  1.5°".  To 
find  the  position  of  the  axis  ab  through  the  center  of  gravity, 
moments  of  the  areas  are  taken  about  the  back  of  the  shorter  leg  as 
follows ; 


Areas,  Levers.            Moments. 

i.S°;;x  25"  =         .375 

3.0°  x  3."  =     9.000 
4.5°"                      9.375 


COMPUTATION  OF  PROPERTIES  OF  SECTIONS. 


Then  dividing  the  total  moment  by  the  total  area,  the  distance  of 
the  centre  of  gravity  from  the  point  of  moments  is  obtained  thus  : 
9,375  -*-  4-5  =  2.08". 

For  the  position  of  the  axis  cd  moments  are  taken  about  the 
back  of  the  longer  flange,  and  divided  by  the  area  as  before. 

Areas.  Levere.          Moments. 


Levere. 

.25"     =         .75 


.2 


3-00 


4-5  D"  3-75 

Then  375  -f-  4.5  =  .833",  which  is  the  distance  from  the  back  of 
longer  leg  to  the  axi^  cd.  The  moment  of  inertia  about  axis  ab 
will  be  computed  as  follows  : 

bh3       .5X63 
I  for  rectangle  (6"  X  .5")  =  —  =  -  =  9.00 

12  12 

bh3       ^X  tj3 
I  for  rectangle  (a"X  .5")  =  —=—  —  =     .03 

12  12 

3.0  D"  X     .Q22=  2.54 
1.5°"  X  1.83'=  5.02 


I  ab         =16.59 


16.59 


/T~          /  16.59 
=  4-23,  rab  =  -*/  —  =  */  --  =  1.92. 

* 


3-92  A  4.5 

The  moment  of  inertia,  moment  of  resistance  and  radius  of  gyra- 
tion about  axis  cd  are  found  similarly. 


«ft]Vcfdf  Wa^e 


*—  7*  — ^ 
Fig.  28 


1 6  BRIDGE  AND  STRUCTURAL  DESIGN. 

Fig.  28  represents  a  24"  I  at  80  Ibs.  Its  moment  of  inertia  about 
the  axis  ab  is  equal  to  that  of  the  circumscribing  rectangle,  less  the 
moment  of  inertia  of  the  voids  about  the  same  axis. 

Area  of  rectangle  7"  X  24"  =  1 68.00 

Area  of  2  rectangles  3.25"  X 22.716"=  141. 15 

3.25  X  .542 
Area  of  4  triangles =  3.52         144.67 


Area  of  beam  —  23.33  SQ-  in. 

MOMENT  OF  INERTIA  ABOUT  AXIS  06. 

bh3  7X243 

For  circumscribing  rectangle  7  X  24  !«&= =  =  8064.  oo 

bh8       /3. 25x21. 7i63\ 
For  2  rectangles  3.25x21.716         \ab= =2 \  ~  j=5547-Oi 

3.25X-542  bh3       /  3.25X-5423  \ 

For  4  triangles       — — Ia&= =41    -  )=        .02 

2  36        V  36          / 

-j-  area  of  triangles  into  square  of  distance  from  axis— 3. 52  X 1 1 -O398  =  428. 95     5975 . 98 

2088.02 

I        2088.02 

=174.00. 


23-33 

MOMENT  OF  INERTIA  ABOUT  AXIS  cd. 

bh8  24X73 

For  circumscribing  rectangle  7x24  Icd=  -  =  -       =  686.00 

bh3       /2i.  716x3.  253\ 
For  2  rectangles  3.25x21.716  -  =2!  -  1=124.24 

-f-  area  of  rectangles  into  square  of  distance  from  axis  141.15x1.875*  =496.  14 

3.25X-542  .    bh8       /    .  542x3.25^ 

For  4  triangles  ^—  -  -  —  -  lcd=  —  =4!  —  ^  -  ^—  —  )=     2.07 

2  36        \  36          / 

-f-  area  of  triangles  into  square  of  distance  from  axis=3.52X2.4i72=  20.56    643.01 

42.99 

I         42.99 

I  #/=  42.99.     R  cd—  —  =  --  =  12.28. 
n  _  3-5 

1  42.99 

=1.36 


A       H  23.33 

It  is  seldom  necessary  to  work  out  the  properties  of  angles, 
beams,  channels,  etc.,  as  they  are  to  be  found  in  the  hand  books 
published  by  the  rolling  mills  companies. 


COMPUTATION  OF  PROPERTIES  OF  SECTIONS. 


The  following  table  gives  the  properties  of  three  special  German 
beams  with  wide  flanges,  used  principally  for  columns.  They  will 
be  referred  to  again  in  the  following  pages : 

Fig.  29.— 


Description 


Area  Wt.        lab  led  Hob  Red  ra&  red 

5.25  17.5  23.22  6.41  9.07  2.81  2.12  i. ii 

7.00  23.3  44.90  II. 01  14.72  4.32  2.55  1.25 

8.60  29.0  96.50  15.50  23.89  5.40  3.35  1.34 


Fig.  30 


Fig.  31 


.  30  represents  a  chord  section  composed  of 


two  15"  [s. 
one  24  X  } 


$  33lbs  =  19.80 
plate     =  12. oo 


31.80°" 

To  find  the  position  of  the  axis  ab  through  the  centre  of  gravity 
of  the  section,  the  simplest  method  is  to  take  moments  of  the  areas 
about  the  centre  line  of  the  channels  and  divide  by  the  total  area. 
The  result  will  be  the  distance  from  the  centre  line  of  channels  to 
the  axis  ab,  thus : 

Areas.  Levers.    Moments. 


Channels 
Plate 

Totals 


19.80°"  X 


o  = 


X7-75  ~ 


o 
93 


93 


then 


93 


31-80 
nel  to  axis  ad. 


=  2.92",  which  is  the  distance  from  centre  line  of  chan- 


18  BRIDGE  AND  STRUCTURAL  DESIGN. 

MOMENT  OP  INERTIA  ABOUT  AXIS  ab. 

bhs       24X-5* 
I  for  24XJ4  plate  about  its  center  of  gravity  =  = =      .25 

Area  of  plate  into  square  of  distance  of  its  centre  of  gravity 

from  axis  ab  =  i2.oX4-838=28o.2o 

Ifor2.i5"[s@33lbsabouttheircentreofgravity(fromCarnegie)=  2x312.6  =625.20 
Area  of  channels  into  square  of  distance  of  their  centre  of 

gravity  from  axis  =i9.8oX2.928=i68.82 

1074.47 

I        1074.47 

00=1074.47.       R  ab— — = =103.1. 

n        10.42 

I  1074-47 

o""^  5'Si- 

31.80 

MOMENT  OF  INERTIA  ABOUT  AXIS  cd. 

bh8      .5x24' 
I  for  24X  %  plate  about  its  centre  of  gravity = —  = =  576.00 

I  fora.  15  [s@33lbs  about  their  centre  of  gravity(fromCarnegie)=  2x8.23=  16.46 
Area  of  channels  into  square  of  distance  of  their  centre  of 

gravity  from  axis  =19. 8X9-  29s =1708. 82 

2301.28 
I          2301.28 

1^=2301.28.    Rr</  = — = =  191.77. 


Fig.  31  represents  a  chord  section  composed  of 

1  cover  plate  24  X  ^  =12.00 

2  web  plates  i6X}4  =16.00 
4  angles  6X3>£X^  =18.00 

46.00°" 

To  find  the  position  of  the  axis  ad  moments  of  the  areas  are  taken 
about  the  centre  line  of  the  webs  thus  : 

Areas          Levers          Moments 

Web  plates  and  angles    34°" X         o  =         o 
Cover  plates  12 a" X    8.25  =       99 

Totals  46°"  99 

99 

then —  =  2.15",  which  is  the  distance  from  centre  line  of  web  plates 
46 

to  axis  ab. 


COMPUTATION  OF  PROPERTIES  OF  SECTIONS.  19 

The  location  of  centre  of  gravity  of  angles  is  obtained   from 
Carnegie. 


MOMENT  OF  INERTIA  ABOUT  AXIS  06. 

bh3      24X  5* 
I   for  24XK   cover  plate   about  us  centre  of  gravity=  -  =  -          =-*5 

bh8         IXI68 
I  for  2.i6x>£  web  plates  about  their  centre  of  gravity=  -  =  -  =  341.34 


I  for  4-6X3^XK  angles  about  their  centre  of  gravity  (from 

Carnegie)  =4X16.59  =  66.36 

Area  of  cover  plate  into  square  of  distance  of  its  centre  of  grav- 
ity from  axis  =12.0X6.10*=  446.52 

Area  of  web  plates  into  square  of  distance  of  their  centre  of 

gravity  from  axis  =16.0X2.15*=  73.96 

Area  of  upper  angles  into  square  of  distance  of  their  centre  of 

gravity  from  axis  =  9-oX3-772=  127.92 

Area  of  lower  angles  into  square  of  distance  of  their  center  of 

gravity  from  axis  =  9.0x8.07*=  586.12 

1642.47 
I          1642.47 

1^=1642.47.     Rad= — = =161.81. 

_n 10.15 

I          /  16,. 

=  5.98. 


MOMENT  OF  INERTIA  ABOUT  AXIS  «L 

bh8        .  5X248 
I   for  24XK  cover  plate  about  its  centre  of  gravity  =  — -  = =  576.00 

bh8       I6X-58 
I  for  2.i6x  Y*  web  plates  about  their  centre  of  gravity  = = =         .37 

I  for  4.6X3>£X>£  angles  about  their  centre  of  gravity  (from 

Carnegie)  =  4X4-25  =  17.00 

Area  of  web  plates  into  square  of  distance  of  their  centre  of 

gravity  from  axis  =i6.ox8.252=io89.oo 

Area  of  angles  into  square  of  distance  of  their  centre  of 

gravity  from  axis  =i8.oX9-338=i566.88 

3249.25 
.  I          3249.25 

I  <:</=  3249.25.     R  cd= — = =270.77 

n 12 

I          /  3249.25 


20 


BRIDGE  AND  STRUCTURAL  DESIGN. 


ART.  11.— EXAMPLES  ILLUSTRATING  THE  METHOD  OF  DETERMINING  THE 
SIZES  OF  BEAMS  REQUIRED  FOR  VARIOUS  CASES.  THE  MAXIMUM 
FIBRE  STRESS  NOT  TO  EXCEED  15,000  LBS.  PER  SQ.  IN. 

Fig:.  32  represents  a  cantilever  beam  with  a  concentrated  load  of 
3,000  Ibs.  10  ft.  from  point  of  support.  M  =  3,000  X  10  =  30,000 
ft.-lbs.  30,000X12=360,000  inch-lbs. 

M      360,000 

R= — = — —24.     Turning:  to  the  table  of  I-beams  in  Car- 

f        15,000 

negfie,  it  will  be  found  that  a  10"  I  @  25  Ibs  has  a  moment  of  resist- 
ance (or  section  modulus)  of  24.4.     This  is  the  beam  required. 


Fi£  36 


Fig:.  33  represents  a  cantilever  beam,  projecting:  12  ft.  from  a  wall, 
and  loaded  with  a  uniform  load  of  500  Ibs.  per  lineal  foot. 

500XI22 

M  —  -  =  36,000  ft.-lbs.   36,000  X  12  =  432,000  inch-lbs. 

12 

432,000 

R  =  ---  =  28.8.    This  case  requires  a  12"  I  @  31.5  Ibs.   R=36. 
15,000 

Fig.  34  represents  a  beam  of  2O-ft.  span,  with  a  centre  load  of 
15,000  Ibs. 

15,000X20 

M  =  ----  =  75,000  ft.  Ibs.     75,000  X  12  =900,000  inch-lbs. 
4 

is  case  requires  a  15"  I  @  45  Ibs.     R=6o.8 


.  15,000 

Fig:.  35  represents  a  beam  of  20  ft.  span,  with  a  uniform  load  of 
600  Ibs.  per  foot. 

600  X2O2 

M  =  ---  =  30,000  ft.-lbs.      30,000  X  12  =  360,000  inch-lbs. 
8 


COLUMNS  AND  STRUTS. 


21 


36o,OOO 

R  = =24.     This  requires  a  10    I  @  25  Ibs.     R=24.4. 

15,000 

Fig-.  36  represents  a  beam  of  20  ft.  span  loaded  unevenly  as  shown. 

For  the  reaction  at  A,  moments  are  taken  about  B,  thus 
4,000  X  5=20,000 
7,000 X  9=63,000 
3,000X16=48,000 

131, ooo  ft.lbs.  then  131,000-^-20=6,550  Ibs.  =  reaction  at  A. 

The  maximum  bending-  moment  will  be  under  the  7,000  Ibs.  load. 
The  reaction  A  acts  with  a  lever  arm  about  this  point  of  n  ft. 
tending:  to  cause  rotation  in  a  right  handed  direction,  and  the  load 


Fig.  37 


Fig.38 


of  3,000  Ibs.  acts  with  a  lever  arm  of  7  ft.  about  the  same  point 
tending  to  cause  rotation  in  a  left  handed  direction,  as  indicated  by 
curved  arrows.  ThenM  =  6,550  Ibs.  X  n'  —  3,000  Ibs.  X  /  =  51,050 

612,600 

ft.-lbs.  51,050  X  12  =  612,600  inch-lbs.  R=  —      —  =  40.8.      This 

15,000 

case  requires  a  15"  I  @  42  Ibs.     R  =  58.9. 

No  beam  should  be  used  where  the  span  exceeds  twenty  times  the 
width  of  the  flange  unless  supported  laterally.  Floor  beams  and 
stringers  in  buildings  are  usually  stayed  at  much  closer  intervals, 
either  by  other  beams  framing  into  them  or  directly  by  the  floor. 

ART.    12.— COLUMNS   AND   STRUTS. 

A  column  or  strut  may  fail  by  crushing,  by  bending  or  by  both 
combined.  A  short  column  will  sustain  a  greater  load  than  a  long 
one  of  the  same  section,  and  a  column  with  fixed  (or  square)  ends, 
more  than  one  with  hinged  (or  pin)  ends.  Columns  are  usually  divi- 


22  BRIDGE  AND  STRUCTURAL  DESIGN. 

ded  into  three  classes,  as  illustrated  in  Figs.  37,  38  and  39.    When 
overloaded,  they  will  bend  as  shown. 

There  are  several  formulas  for  proportioning  columns,  but  Ran- 
kine's  (or  Gordon's)  is  the  one  in  most  general  use,  and  agrees 
closely  with  experiments.  In  it,  a  certain  permissible  unit  stress 
for  direct  crushing  is  assumed,  which  is  reduced  for  each  special 
case,  depending  on  the  length  of  column,  its  radius  of  gyration,  and 
the  condition  of  its  ends,  whether  square  or  pin. 


RANKINE'S   FORMULA. 
For  square  ends  P= 


36,000  rg 
F 


One  pin  and  one  square  end      P=  I 


Both  pin  ends  P= 


24,000  r* 

F 
I2 
18,000  r2 


In  which  F  =  permissible  compression  per  square  inch  for  direct 

crushing. 

P  =  permissible  compression  per  square  inch  for  col- 
umns. 

1  =  length  in  inches, 
r  =  least  radius  of  gyration  in  inches. 

To  find  the  permissible  stress  per  square  inch  for  a  column  it  is 
necessary  to  know  its  radius  of  gyration ;  so  some  section  must  be 
assumed,  and  if  found  to  be  too  small  or  too  large  it  will  be  nec- 
essary to  try  another.  The  work  is  greatly  simplified  by  using  the 
following  table,  in  which  F  =  12,000  Ibs.,  1  =  length  of  column  in 
feet,  and  r  =  least  radius  of  gyration  in  inches.  For  convenience, 
the  length  is  taken  in  feet,  and  is  multiplied  by  the  factor  12  in  the 
formula  to  reduce  it  to  inches. 


COLUMNS  AND  STRUTS. 


PERMISSIBLE  COMPRESSION   PER  SQ.  INCH  FOR  COLUMNS 

12,000  LBS.  REDUCED  BY  RANKINE'S  FORMULA 
Square  Ends  Pin  and  Square  Ends  Pin  Ends 

I2OOO  I2OOO  I2OOOO 


36000  r* 


24000  r' 


18000 


1 

Square 

Square  and 

Pin 

l 

Square 

Square  and 

Pin 

r 

Ends 

Pin  Ends 

Ends 

r 

Ends 

Pin  Ends 

Ends 

2.0 

11,820 

11,720 

11,630 

7.6 

9,750 

8,910 

8,210 

2.2 

11,780 

11,660 

".550 

7-8 

9,650 

8,790 

8,070 

2.4 

H,730 

11,600 

11,470 

8.0 

9-560 

8,670 

7-940 

2.6 

II,  680 

",530 

11,390 

8.2 

9,46o 

8,550 

7,800 

2.8 

11,640 

11,460 

11,290 

8.4 

9,36o 

8,430 

7,670 

3-o 

11,580 

n,390 

11,190 

8.6 

9,260 

8,310 

7,540 

3-2 

11.530 

11,310 

11,090 

8.8 

9,l6o 

8,190 

7,410 

3-4 

11,470 

11,220 

10,  980 

9.0 

9,060 

8,080 

7,280 

3-6 

11,410 

11,130 

10,870 

9-2 

8,970 

7,96o 

7,i6o 

3-8 

11.350 

11,040 

io,  760 

9.4 

8,870 

7,840 

7.030 

4-0 

11,280 

10,950 

10,640 

9-6 

8,770 

7,730 

6,910 

4.2 

II.2IO 

10,850 

10,520 

9-8 

8,670 

7,6lO 

6,790 

4-4 

11,140 

10,750 

10,390 

IO.O 

8,570 

7,500 

6,670 

4.6 

II,  060 

10,650 

10,260 

10.2 

8,480 

7,390 

6,550 

4-8 

10,990 

10,540 

10,130 

10.4 

8,380 

7,280 

6,430 

5-0 

10,910 

10,440 

10,000 

10.6 

8,280 

7,170 

6,320 

5-2 

10,830 

10,330 

9,870 

10.8 

8,180 

7,060 

6,210 

5-4 

10,750 

10,220 

9,730 

II.O 

8,090 

6,950 

6,100 

5-6 

10,660 

10,100 

9,600 

II.  2 

7,990 

6,850 

5,990 

5-8 

10,580 

9,990 

6,460 

II-4 

7,900 

6,740 

5,880 

6.0 

10,490 

9,870 

9,320 

u.6 

7,800 

6,640 

5,78o 

6.2 

10,400 

9.75° 

9,180 

11.8 

7,710 

6,540 

5,68o 

6.4 

IO,3IO 

9.630 

9,040 

12.0 

7,620 

6,440 

5,58o 

6.6 

10,220 

9.5io 

8,900 

12,2 

7,520 

6,340 

5,48o 

6.8 

10,  130 

9,390 

8,760 

12.4 

7,430 

6,240 

5,38o 

7.0 

10,030 

9,270 

8,620 

12.6 

7,340 

6,150 

5,290 

7.2 

9,940 

9,i5o 

8,480 

12.8 

7,250 

6,060 

5,190 

7-4 

9,840 

9,030 

8,350 

13.0 

7,160 

5,960 

5,100 

Supposing  it  is  required  to  know  the  permissible  stress  per  square 
inch  on  a  column  20  feet  long,  with  two  pin  ends  and  radius  of 
gyration  of  4  inches, 

Then — = — =5-     In  the  column  headed  "Pin  Ends,"  and  op- 

r    4  i 

posite  the  number  5.0  in  column  for  — ,  will  be  found  10,000,  which 

is  the  unit  stress  required. 


24  BRIDGE  AND  STRUCTURAL  DESIGN. 

The  same  table  may  be  used  for  any  other  value  of  F  (say  15,000), 
for  example:  If  10,000  Ibs.  per  square  inch  be  the  permissible  stress 

when  the  factor  F=  12,000,  then  io,oooX  — =  12,500   Ibs.    per 

12,000 

square  inch  is  the  permissible  stress  when  the  factor  F  =  15,000. 

Columns  in  building  are  usually  figured  as  though  they  had  pin 
ends,  which  is  an  error  on  the  safe  side  and  compensates  to  a  cer- 
tain extent  for  eccentric  loading  and  any  slight  unevenness  in  the 
foundation.  The  end  sections  of  top  chords  in  pin-connected 
bridges  are  usually  figured  as  "square  and  pin  end"  columns,  and 
the  intermediate  sections  as  square  end  columns.  The  posts  are 
usually  figured  as  "pin  end"  columns  whether  connected  by  pins 
or  otherwise.  These  rules  are  not  always  adhered  to  and  one  must 
be  guided  by  the  specification  under  which  the  structure  is  to  be 
designed. 

ART.   13.— EXAMPLES   ILLUSTRATING  METHOD   OP  DESIGNING   COLUMNS 

AND    STRUTS. 

A  column  20  ft.  long  is  required  to  support  a  load  of  150,000  Ibs. 
Unit  stress,  12,000  Ibs.,  reduced  for  pin  ends  by  Rankine's  formula. 


Fig.  40  represents  a  trial  section  consisting  of  : 

i—  6^X5^  XYs  I  @  23.3  Ibs.  =  7.00     (Table  Art.  10.) 
2  —  10"  [s  @  15  Ibs.  8.92     (Carnegie) 

15.92  square  inches. 
For  the  least  radius  of  gyration  : 

I  ab=(ior  2.io"[s.     See  Carnegie)     66.9X2=133.80 
(for  6^X5^  I.     See  table  Art.  i  o)  n.oi 

144.81 


/"T 

=  <\  —  = 
* 


lab  =  <\  —  =  A/  -  =  3-02 
15.90 


RIVETS  AND  RIVETING.  2$ 

I  cd  (for  6>6X5>6  I.     See  table  Art.  10)  =  44.90 

(for  2.io"[s  about  c.  g.     See  Carnegie)  2.30X2=  4.60 

Area  of  [s.  into  sq.  of  dist.  of  their  e.g.  from  ^=8.92X3. 72  =  122.50 

/  I  /I72.00  172.00 

r  cd  =   V  ~T  =  A/  "  =  3-28 

"  A         '  15.92 

1  20 

Thus  the  least  radius  of  gyration  =  3.02",  and  —  = =  6.6. 

r          3.02 

The  value  corresponding  to  this  in  table  (Art.  12)  is  8,900.  Then 
150,000  -f-  8,900  =  16.8  sq.  inches  required.  The  trial  section  is 
slightly  too  small,  but  it  is  only  necessary  to  use  heavier  channels  of 
the  same  size.  The  radius  of  gyration  will  remain  practically  the 
same.  The  column  should  be  made  of 


1—6^  X  5^  X  Y*  I  @  23.3  Ibs.  =    7.00 

2 — IO"[S  @  20  Ibs.  =  11.76 

18.76  sq.  inches 
Columns   sustaining   comparatively   light   loads   are   frequently 

made  of  a  single  I-beam  with  wide  flanges   (see  table,  Art.  10). 
A  column  16  ft.  long  is  required  to  sustain  a  load  of  45,000  Ibs. 

Unit  stress  12,000  Ibs.,  reduced  for  pin  ends  by  Rankine's  formula. 
The  area  of  a  6J  x  5^  x  £  I  is  7.0  sq.  ins.,  and  the  least  radius  of 

1  16 

gyration,  1.25".  —  = =12.8.    The  value  corresponding  to  this 

r         1.25 

in  table  (Art.  12)  is  5,190  Ibs.  per  sq.  in.    Therefore  the  capacity  of 
this  column  is  5,190  x  7.0  sq.  ins  =  36,330  Ibs.,  which  is  too  small. 
The  area  of  an  8  X  sJ  X  TV  I  is  8.6  sq.  ins.     The  least  radius  of 
1  16 

gyration  is  1.34".  —  = =  11.9.    The  value  corresponding  to 

r  1-34 

this  in  table  is  5,630  Ibs.  per  sq.  in.     Then  5,630  x  8.6  sq.  ins.  = 
43,500  Ibs.    Therefore  this  beam  is  suitable  for  the  purpose. 

ART.   14.      RIVETS  AND  RIVETING. 

Sizes  of  Rivets  used  in  structural  steel  work  are  J",  f ",  f ",  }"  and 
i",  Those  in  most  general  use  are  J  and  f .  The  smaller  ones  are 
used  only  in  very  light  members  to  avoid  cutting  out  too  much  sec- 
tion. Rivets  larger  than  J  are  only  used -when  it  is  impossible  to 
get  in  enough  of  a  smaller  size  for  lack  of  space. 


26  BRIDGE  AND  STRUCTURAL  DESIGN. 

Spacing  of  Rivets. — The  distance  center  to  center  of  rivets  should 
not  be  less  than  three  times  their  diameter.  In  compression  mem- 
bers, rivets  should  not  be  further  apart  than  sixteen  times  the  thick- 
ness of  the  outside  plates,  in  line  of  stress,  and  generally  should 
not  exceed  six  inches.  The  distance  from  centre  of  rivet  to  end  of 
member  should  not  be  less  than  one  and  one-half  times  the  diameter 
of  rivet  or  generally  I \  ins.  for  f  rivets,  and  i£  ins.  for  J  rivets. 

Size  of  Rivet  Holes. — In  ordinary  work  the  holes  are  punched 
1-16"  larger  than  rivet,  but  in  more  particular  work  they  are 
punched  about  J"  smaller,  and  reamed  after  assembling  to  1-16" 
larger  than  rivet.  Holes  cannot  be  punched  in  metal  of  greater 
thickness  than  diameter  of  rivet,  and  in  this  case  they  must  be 
drilled.  In  tension  members,  the  area  of  the  rivet  holes  is  deducted 
from  the  gross  area,  allowance  being  made  for  holes  \"  larger  than 
rivets.  In  compression  members  no  allowance  need  be  made  for 
rivet  holes. 

Strength  of  Rivets. — Rivets  may  fail  by  shearing,  by  crushing 
(or  bearing)  or  by  tension  on  the  heads.  Generally  it  is  not  con- 
sidered good  practice  to  use  rivets  in  tension,  as  their  strength  in 
this  direction  is  somewhat  uncertain  on  account  of  the  initial  stress 
in  them  from  cooling,  but  it  is  sometimes  unavoidable  to  use  rivets 
in  this  way. 

Permissible  Unit  Stresses. — In  buildings  and  highway  bridges 
9,000  Ibs.  per  square  inch  for  shear  and  18,000  Ibs.  per  square  inch 
for  bearing  are  usually  allowed  for  shop  driven  rivets ;  and  for  field 
rivets,  7,500  Ibs.  shear  and  15,000  bearing. 

In  railway  work  7,500  Ibs.  per  square  inch  shear  and  15,000  Ibs. 
per  square  inch  bearing  are  the  usual  unit  stresses  for  shop  rivets ; 
and  6,000  Ibs.  shear  and  12,000  Ibs.  bearing  for  field  rivets. 

Shearing  and  Bearing  Value  of  Rivets. — The  shearing  value  of  a 
rivet  is  equal  to  the  area  of  its  cross  section  multiplied  by  the  per- 
missible shear  per  square  inch.  Thus  the  shearing  value  of  a  j" 
rivet  at  7,500  Ibs.  per  square  inch  =  .4418  square  inches  x  7,500  Ibs. 
=3,310  Ibs.  The  bearing  value  is  equal  to  the  diameter  of  the  rivet 
multiplied  by  the  thickness  of  metal  on  which  it  bears,  multiplied  by 
the  permissible  bearing  per  square  inch.  Thus  the  bearing  value  of 
a  |  rivet  on  a  f"  plate  at  15,000  Ibs  per  square  inch  =  f  x  f  x  15,000 
Ibs.  =  4,220  Ibs. 

Rivets  may  be  either  in  single  or  double  shear.  In  this  first  case 
the  joint  could  fail  by  the  rivets  shearing  in  one  plane  only,  that  be- 


THE  COMPLETE  DESIGN  OF  A  ROOF  TRUSS.  27 

tween  the  two  members  joined  (see  Fig.  41).  When  rivets  are  in 
double  shear,  they  would  have  to  be  sheared  in  two  planes,  as  shown 
in  Fig.  42,  before  the  joint  could  fail  in  that  manner,  and  they  would 
have  twice  the  value  of  rivets  in  single  shear.  In  most  cases,  how- 
ever, when  rivets  are  in  double  shear,  their  bearing  value  is  less 


/""N     .. /"""N       plane  of  sS-iesrinft  /*"~"\       /"~"N      ,e!anto{  staaiwfc 

f£-S        j    j    i/yj. „  f       k     .  W       I,   !    l/r  , — r 


Fig.  *H  Fig  4-2 

than  twice  their  shearing  value,  and  so  the  bearing  value  determines 
the  strength  of  the  joint.  In  Fig.  41  each  rivet  is  good  for  3,310  Ibs. 
in  shear  and  4,220  Ibs.  in  bearing,  so  the  shearing  value  governs. 
In  Fig.  42  the  rivets  are  each  good  for  3,310  x  2  =  6,620  Ibs.  in 
shear  and  4,220  Ibs.  in  bearing,  therefore  the  bearing  value  deter- 
mines the  strength.  If  the  centre  member  were  \"  thick  the  bearing 
value  of  each  rivet  would  be  J  x  -£  x  15,000  =  5,620  Ibs. 

In  designing  a  riveted  connection  great  care  must  be  taken  al- 
ways to  use  the  least  value  a  rivet  can  have  under  the  circumstances, 
whether  single  shear,  double  shear  or  bearing, 

Tables  of  shearing  and  bearing  values  of  rivets  may  be  found  in 
Carnegie's,  Pencoyd's  and  other  hand-books. 

CONVENTIONAL    SIGNS    FOB  -EIVETING     IN    GENERAL     USE     IN    CANADA    AND    THE 

UNITED  STATES. 

Is  ill  aSi  111  **-  ^—  **-  ^  *5   5s 


Shop  Rivets  00a8i 

Field  Rivets  ©       ©      S5      !S 

ART.  15.  THE  COMPLETE  DESIGN  OF  A  ROOF  TRUSS  FOR  BUILDING 
WITH  MASONRY  OR  BRICK  WALLS.  CAPABLE  OF  WITHSTANDING 
WIND  PRESSURE. 

Data :  Width  of  building  out  to  out  of  walls  40'. 
Thickness  of  Walls,  i'  6". 
Span  of  trusses  38'  6"  centre  to  centre  of  bearings. 


28  BRIDGE  AND  STRUCTURAL  DESIGN. 

Slope  of  rafters  30°. 

Trusses   spaced   16'   centres. 

Total  load,  including  weight  of  trusses,  roof  covering,  snow  and 

wind  50  Ibs.  per  square  foot  of  horizontal  projection. 

Unit  Stresses:  Tension,  15,000  Ibs.  per  square  inch. 

Compression,  12,000  Ibs.  per  square  inch,  reduced  by  Rankin's 
formula,  the  rafters  to  be  considered  as  columns  with  square 
ends,  and  the  compression  web  members  as  columns  with  pin 
ends,  length  not  to  exceed  120  times  least  radius  of  gyration. 
Rivet  shear,  7,500  Ibs.  per  square  inch. 
Rivet  bearing,  15,000  Ibs.  per  square  inch. 

Total  load  on  truss  =  38.s'Xi6'X5o  Ibs.  =  30,800  Ibs.,  and 
since  the  rafters  are  divided  into  8  panels,  each  panel  load  = 
30,800  -*-  8  =  3,850  Ibs. 

4,850X16 

Purlins.— Span    16 ,    load    3,850    Ibs.     M  =  =  7,700 

8 

ft.-lbs  =  92,400  inch-lbs. 

M     92,400 

R  = —  = =  6.16.  For  intermediate  purlins  6'  I  s  @  12.25 

f       15,000 

Ibs.  will  be  used,  for  which  R  =  7.3.  For  the  end  purlins  which 
support  only  one-half  panel  load  and,  at  the  centre  where  there 
are  two  purlins,  6"  [  s  @  8  Ibs.  will  be  used. 

The  truss  is  of  the  Fink  pattern,  and  the  method  of  constructing 
the  stress  diagram  is  fully  described  in  Art.  3.  The  stresses,  which 
are  scaled  from  the  stress  diagram  Fig.  433,  are  all  written  on  the 
diagram  of  truss,  Fig.  43.  The  required  areas  of  the  tension  mem- 
bers are  obtained  directly  by  dividing  the  stresses  by  the  permissible 
unit  stress  as  shown.  Then  suitable  angles  are  selected  from  the 
hand-books  of  the  rolling  mills  which  give  the  areas  for  all  standard 
sizes.  Allowance  must  be  made  for  rivet  holes :  in  members  sub- 
jected to  small  stress,  and  connected  by  one  leg  only,  the  area  of  but 
one  hole  need  be  deducted  from  each  angle ;  but  where  angles  are 
connected  by  both  legs  it  is  advisable  to  allow  for  two  holes  in  each 
angle.  Angles  requiring  more  than  three  rivets  should,  when  pos- 
sible, be  connected  by  both  legs.  In  the  present  example  f "  rivets 
will  be  used,  and  allowance  made  for  J"  holes.  No  angle  smaller 
than  2.\  x  2  x  \  will  be  used,  and,  when  necessary  to  connect  both 
legs,  2j  x  2j  x  J  will  be  the  minimum. 


THE  COMPLETE  DESIGN  OF  A  ROOF  TRUSS. 


Fig.  433. 


Referring1  to  diagram  Fig.  43,  it  will  be  seen  that  member  a O 
requires  1.57  °' ; 

then  two  3X2>£xXLs  =  2.62  °"  gross  area 

less  4*  7/8  X  %          =  Jfy  v"  area  of  4  holes 

1.75  D"net  area. 

In  order  not  to  make  too  many  splices,  the  same  angles  will  be 
be  used  for  member  b^O. 
Member  d^O  requires  .89°' . 

Two  3Jfj$a]4  X  #  Ls  =  2.38  n"  gross  area 
less  4  X  7/8  X  %        —    .87  a"  area  of  4  holes 

1.51  D"  net  area. 


30  BRIDGE  AND  STRUCTURAL  DESIGN. 

Member  dd^  requires  .67  D".  Since  the  stress  in  this  member  is 
small  it  only  requires  to  be  connected  by  one  flange. 

Two  2%  X  2  X  Y^  Ls  —  2.12  °"  gross  area 
less    2  X  ^3  X  ^         =     .44°"  area  of  2  holes 

1.68  °"  net  area. 

The  same  angles  will  be  used  for  members  c^. 
Members  bb^  and  ccl  each  require  .22  °". 

One  2*/2  X  2  X  %  L  =  1.06  a"  gross  area 
less  i  X  fa  X  ^       =     .22  °"  area  of  i  hole 

.84  D  "  net  area. 

Members  ab  and  cd  each  sustain  a  compressive  stress  of  3,300  Ibs. 
and  their  length  is  3.2  feet  (about).  One  2>£  X  2  X  X  L  is  assumed 

1       3.2 
and  its    least  r  =  .43.     — =  7«4-     By  table  of  compression 

r       43 

values  (Art.  12),  the  permissible  stress  per  square  inch  =  8,350  Ibs., 
then  3, 300 -r-  8,350  =  .40  D"  required,  and  the  area  of  one  2.^/2,  X 
2  X  %  L  =  i. 06  D".  Rivet  holes  are  not  deducted  from  the  area 
of  compression  members,  as  the  rivets  are  supposed  to  fill  the  holes 
completely,  and  transmit  the  pressure  from  one  side  of  the  hole  to 
the  other. 

Member  b,cl  sustains  a  compressive  stress  of  6,700  Ibs.  and  its 
length  is  about  6.4  ft.  Two  2^2  X  2  X  ^  Ls  are  assumed,  with 
the  longer  legs  back  to  back,  but  separated  about  X",  thus:  1  f,  to 
straddle  the  connection  plates  at  the  ends.  Tables  of  radii  of  gyra- 
tion for  two  angles  are  given  in  Carnegie's  and  Pencoyd's  hand 
books.  They  are  computed  for  the  maximum  and  minimum  thick- 
nesses only,  but  values  for  intermediate  thicknesses  may  be  inter- 
polated with  sufficient  accuracy.  From  these  tables  the  least  radius 
of  gyration  for  two  2^/2  X  2  X  ^  Ls  as  above  is  found  to  be  .80". 

1       6.4 

Then — =  —  =8  which  (in  table  Art.  12)  corresponds  to  a  unit  stress 
r       .80 

of  7,040  Ibs.  and  6,700  -*-  7,940  =  .84  square  inches  required.  The 
area  of  two  2^  X  2  X  #  Ls  —  2.12  D"  which  is  more  than  twice 
the  area  required,  but  a  single  angle  would  be  too  small,  because 
its  radius  of  gyration  would  be  much  less. 

Member  Aa  sustains  a  compressive  stress  of  27,000  Ibs.  and  its 
length  is  about  5.5  feet.  Two  3  X  2%  X  ^  Ls  are  assumed,  with 


THE  COMPLETE  DESIGN  OF  A  JROOF*  TRUSS. 


32  BRIDGE  AND  STRUCTURAL  DESIGN. 

the   longer  legs  back  to  back,  thus  1  f.     The  area  =  2.62  °  "  and 

the  least  r  =  .95.     Then  —  =  —  =  5.8,    which  corresponds  to   a 

r       -95 

permissible  unit  stress  of  10,580  Ibs.  for  square  ends,  and  27,000  -*• 
10,580  =  2.55  n  "  required.  Therefore  the  trial  section  is  suitable, 
its  area  being  slightly  greater  than  that  required. 

It  is  unnecessary  to  consider  the  remaining  panels  of  the  rafters 
as  the  same  angles  will  be  used  throughout. 

At  the  centre  of  the  truss  a  small  angle  hanger  is  used  to  prevent 
the  bottom  chord  from  sagging. 

Details. — Fig.  44  is  a  detail  drawing.  The  height  at  centre  is  ob- 
tained by  multiplying  one-half  the  centre  to  centre  span  by  the  tan- 
gent of  30°. 

At  the  ends  of  truss  a  f "  plate  is  used  to  connect  the  rafters  with 
the  bottom  chord.  The  rivets  are  in  double  shear.  Referring  to 
table  of  rivet  values  in  Carnegie  or  some  other  hand-book,  the 
shearing  value  of  f  rivets  at  7,500  Ibs  =  2  x  3,310  =  6,620  Ibs.,  but 
the  bearing  value  at  15,000  is  only  4,220  Ibs.,  which  latter  must  be 
used. 

The  number  of  rivets  required  in  rafter  =  27,000  -f-  4,220  =  7 ; 
and  the  number  of  rivets  in  bottom  chord  =  23,500  -f-  4,220  =  6. 
4  rivets  are  used  in  the  main  angles  of  bottom  chord  and  2  rivets  in 
the  lock  angles.  This  arrangement  not  only  requires  a  smaller  gus- 
set plate  than  if  all  rivets  were  put  in  one  line,  but  it  distributes  the 
stress  much  better  in  the  angles. 

The  bearing  plates  on  the  walls  must  be  large  enough  to  dis- 
tribute the  load.  If  the  trusses  are  to  rest  directly  on  a  brick  wall, 
the  load  per  square  inch  should  not  exceed  100  Ibs.  Since  the  total 
load  on  truss  is  30,800  Ibs.  the  reaction  at  each  end  will  be  30,800  x 
£  =  15400,  and  15,400  -f-  1 00  =  154  square  inches  required  in  bear- 
ing plate.  The  plate  used  (QX  18=  162  square  inches)  slightly  ex- 
ceeds this  area. 

In  member  ad  there  is  a  stress  of  3,300  Ibs.  The  rivets  in  this 
case  are  in  single  shear  =  3,310  Ibs.,  but  the  bearing  value  on  #'' 
plate  is  only  2,810  Ibs.  Two  rivets  are  sufficient  for  this  member, 
as  well  as  for  cdt  ddl  and  cct. 

In  member  dlcl  there  is  a  stress  of  6,700  Ibs.  The  rivets  are  in 
double  shear  and  their  bearing  value  on  TV  plate  =  3,520  Ibs.  each. 
Two  rivets  will  do  here  also. 

The  bottom  chord  is  spliced  at  panel  points  near  centre  of  span. 


ROOF  TRUSSES  SUPPORTED  BY  STEEL  COLUMNS.  33 

In  b^O  the  stress  =  20,000  Ibs.  and  the  value  of  the  connection  is 
as  follows: 

3  rivets  in  bearing-  on          ^6  plate  @  4,220  Ibs.  =  12,660 

4  rivets  in  bearing-  on  $X><  }i  plate  @  2,810  Ibs,  =  11,240 

23,900  Ibs. 

In  d^O  the  stress  =  13,000  Ibs.  and  the  value  of  connection  is  : 
2  rivets  in  bearing:  on  y&  plate  @  4,220  Ibs.  =    8,440 

4  rivets  in  bearing  on  5>£X^  plate  @  2,810  Ibs.  =  11,240 

19,680  Ibs. 

At  apex  of  rafters  there  is  a  $6  giisset  plate  and  the  rivets  are 
g-ood  for  4,220  Ibs.  each. 

The  stress  in  Dd=2i,ooo,  then  21,000-^-4,220=5  rivets  required. 

The  stress  in  ddl  =10,000,  then  10,000-^4,220=3  rivets  required. 

The  gusset  plates  at  ends  of  truss  and  at  apex  extend  above  the 
rafters.  By  this  arrangement  the  stresses  in  them  are  better  dis- 
tributed. 

Purlins. — Although  the  purlins  are  designed  for  vertical  loads, 
for  convenience  they  are  set  normal  to  the  rafters.  If  unsupported 
laterally  they  would  be  liable  to  fail  through  side  bending,  but  the 
roof  covering  is  depended  on  for  this  contingency. 

ART.    16.— ROOF    TRUSSES    SUPPORTED    BY    STEEL    COLUMNS. 

When  roof  trusses  are  set  on  solid  brick  or  masonry  walls  having 
sufficient  stability  to  withstand  the  wind  pressure,  it  is  not  usually 
customary  to  figure  the  wind  stresses  in  the  trusses  separately,  the 
vertical  load  being  assumed  large  enough  to  cover  everything,  as  in 
Art.  15.  But  when  supported  by  steel  columns  and  braced  thereto 
to  resist  the  overturning  effect  of  the  wind,  it  is  advisable  to  treat 
the  wind  force  and  the  vertical  loads  separately.  Wind  is  usually 
taken  at  30  Ibs.  per  square  foot  acting  in  a  horizontal  direction 
against  a  vertical  plane,  and  on  sloping  surfaces  it  is  reduced  by 
the  following  table  of  co-efficients  which  are  based  on  Unwin's  ex- 
periments. 

CO-EFFICIENTS  FOE  WIND  PBESSUBE  NOEMA.L  TO  PLANE  OF  EOOF. 


00000 


Angle  of  Roof       5          10         20         30         40         50     60   to  90 
Co-efficient         .125       .24        .45        .66        .83        .95          i.oo 

The  vertical  load  consisting  of  the  weight  of  trusses,  roof  cover- 
ing and  snow  may  then  be  taken  at  about  35  Ibs.  per  square  foot  of 
horizontal  projection. 


34  BRIDGE  AND  STRUCTURAL  DESIGN. 

AET.  17.-THE  DESIGN  OF  A  KNEE-BEACED  MILL  BUILDING. 

Data  :  Width  of  building-  40'  o"  centre  to  centre  of  posts. 
Height  of  posts  18'  o" '. 
Angle  of  roof  30°. 

Trusses  spaced  16'  o"  centre  to  centre. 
Roof  covered  with  3"  X  5"  planks  on  edge. 
Sides  covered  with  3"  tongued  and  grooved  planks,  fastened 

to  posts  with  3"  railway  spikes. 

Roof  load  (dead  load  and  snow)  40  Ibs.  per  sq.  ft.  of  hori- 
zontal projection. 

Horizontal  wind  force  30  Ibs.  per  sq.  ft. 
Wind  pressure  normal  to  roof.     30  Ibs.  X  .66  —  20  Ibs.  per 

sq.  ft. 

Unit  stresses  same  as  in  Art.  15. 

Wind  Stresses. — The  wind  pressure  on  side  of  building,  Fig.  45, 
is  assumed  to  be  concentrated  at  top  of  post,  at  foot  of  knee  brace, 
and  at  base  of  post.  The  last  is  neglected,  as  it  has  no  overturning 
effect  on  building. 

Horizontal  wind  force  at  top  of  post  =  16'  X  — '—  X  30  Ibs.  =  1,200  Ibs. 

"     "  foot  of  knee  brace  =  16'  X  —  X  30  Ibs.  =  4,320  Ibs. 

Wind  pressure  on  roof  =  23'  X  16'  X  20  Ibs.  =  7,360  Ibs.,  say  7,400 
Ibs.  The  intermediate  panel  loads  will  then  be  7,400  -*-  4  =  1 ,850  Ibs. 
each,  and  the  end  panel  loads  925  Ibs.  each,  as  shown  on  diagram. 
The  resultant  of  the  wind  on  roof  acts  at  the  middle  point  of  rafter. 
Its  vertical  component  =  6,430  Ibs.  and  its  horizontal  component 
=  3,700  Ibs. 

Reactions. — The  horizontal  forces  are  assumed  to  be  resisted 
equally  by  both  posts,  which  assumption  is  undoubtedly  accurate 
enough  for  all  practical  purposes. 

Horizontal  reaction  for  each  post  =  (3,700  +  1,200  +  4,320)  X  Yz  =  4,6io  Ibs. 

If  the  posts  were  free  to  rotate  at  their  base,  the  vertical  reactions 
due  to  the  horizontal  forces  would  be  obtained  by  taking  moments 
of  these  forces  about  foot  of  posts  and  dividing  by  their  distance 
centre  to  centre.  But  the  posts  are  more  or  less  fixed  by  the  dead 
load  of  roof  and  walls,  also  by  the  anchor  bolts,  if  properly  built 
into  foundations.  Consequently  there  will  be  a  point  of  no  moment 
somewhere  between  base  of  posts  and  foot  of  knee  brace.  This 
point  of  no  moment,  or  of  contra-flexure,  should  never  be  assumed 


THE  DESIGN  OF  A  KNEE-BRACED  BUfLDING. 


35 


higher  than  half-way  between  base  of  post  and  knee-brace  con- 
nection. The  existence  of  a  resisting:  moment  at  foot  of  each  post 
changes  the  vertical  reactions  from  those  determined  by  pure  statics. 
Taking1  the  weight  of  roof  at  20  Ibs.  per  sq.  ft.  and  of  the  sides  at 
10  Ibs.  per  sq.  ft.  the  dead  load  on  post  is  as  follows  : 

Roof  1 6'  X  20'  X  20  Ibs.  =  6,400 
Side    1 6'  X  18'  X  10    "     =  2,880 

Total 9,280  Ibs. 

This  force  is  assumed  to  act  at  centre  of  post  which  is  taken 


F\g.  45 

12"  wide,  with  a  base  plate  20"  wide.  It  will  then  have  a  lever  arm 
of  10"  about  edge  of  base.  One-inch  anchor  bolts  are  assumed, 
which,  allowing  for  thread,  are  equivalent  to  It"  dia.  =  .52  sq.  in. 
each.  The  value  of  one  bolt  will  be  .52  sq.  in.  X  15,000  Ibs.  =  7,800 
Ibs.,  and  it  will  have  a  lever  arm  of  18"  from  edge  of  base. 

Then,  moment  of  resistance  at  base  =  9,280  Ibs.  X  10"  =    92,800 

7,800    "    X  1 8"  =  140,400 

233,200  in. -Ibs. 


36  BRIDGE  AND  STRUCTURAL  DESIGN. 

The  horizontal  reaction  multiplied  by  the  distance  from  base  of 
post  to  point  of  contra-flexure  will  be  equal  to  the  moment  of 
resistance  at  base ;  therefore,  distance  to  point  of  contra-flexure  = 

2^^  2OO 

- — 7 — •  =  50  inches.     The  plane  of  contra-flexure  will  be  assumed 
4,010 

4  ft.  above  base,  where  the  posts  are  considered  to  be  hinged,  as 
shown. 

For  the  vertical  reactions  due  to  horizontal  wind  forces,  moments 
of  these  forces  will  be  taken  about  the  hinges. 

VERTICAL  REACTION   LA. 

3,700  X  19-75  =    73.075 

1,200  X    14          =      16,800 

4,320  X    9       =    38-880 

From  Horizontal  Wind  forces  =  128,755  ft.-lbs.  -f-  40'  =  —  3,220 
"      Vertical  Wind  forces      =  6,430  X  $£  =  +  4,820 

+  z,6oo  Ibs. 
VERTICAL  REACTION   J  K. 

From  Horizontal  Wind  forces  =  +  3,220 

Vertical  Wind  forces  6,430  X  #  =  +  1,610 

+  4,830  Ibs. 

Stress  Diagram.  In  order  to  proceed  with  the  stress  diagram 
without  further  figuring,  imaginary  struts  are  provided,  as  shown 
in  dotted  lines. 

The  external  forces  may  now  be  laid  off  and  the  stress  diagram 
constructed  as  follows  :  Beginning  with  the  force  AB,  the  external 
forces  are  taken  in  regular  order  in  going  around  the  frame  in  a 
right-handed  circular  direction,  and  plotted  in  the  stress  diagram, 
the  last  force  LA  closing  the  diagram.  These  external  forces  are 
shown  in  heavy  lines.  For  wind  stresses  it  is  necessary  to  construct 
the  stress  diagram  for  the  whole  truss,  as  the  stresses  on  opposite 
sides  are  very  different.  Beginning  at  the  left  hand  hinge,  there 
are  two  known  forces  KL  and  LA,  and  two  unknown  forces  Aa  and 
aK.  From  the  point  A  in  diagram  of  external  forces  a  line  is 
drawn  parallel  with  the  member  Aa;  and  from  the  point  K,  a  line 
parallel  with  aK,  the  two  lines  intersecting  in  the  point  a.  In  going 
around  this  joint  in  a  right  handed  direction,  and  following  the 
forces  in  the  stress  diagram,  it  will  be  observed  that  Aa  acts 
towards  the  joint,  which  indicates  that  the  member  is  in  compres- 
sion; and  that  aK  acts  away  from  it,  which  indicates  tension. 


DESIGN  OF  A  KNEE-BRACED  BUILpING.  37 

Next,  at  foot  of  knee  brace,  there  are  now  but  two  unknown  forces 
Bb  and  ba.  From  the  point  B  in  stress  diagram,  a  line  is  drawn 
parallel  with  the  member  Bb; .  and  from  the  point  «,  a  line  parallel 
with  ba,  the  two  intersecting-  in  the  point  b.  Bb  acts  towards  the 
panel  point  under  consideration,  indicating  compression ;  and  ba 
towards  it  also  indicating  compression.  At  top  of  post  there  are 
now  only  two  unknown  forces,  Dd  and  db.  From  point  D  in  stress 
diagram  a  line  is  drawn  parallel  with  Dd;  and  from  the  point  by  a 
line  parallel  with  db,  intersecting  in  the  point  d.  Dd  acts  towards 
the  panel  point,  indicating  compression  ;  and  db  away  from  it  in- 
dicating tension.  The  next  joint  to  be  considered  is  panel  point 
DE.  From  point  E  in  stress  diagram,  a  line  is  drawn  parallel 
with  member  Ee;  and  from  point  d,  a  line  parallel  with  member 
de,  the  two  intersecting  in  point  e.  Ee  acts  towards  the  joint,  in- 
dicating compression  ;  and  ed  towards  it,  indicating  compression. 
At  the  point  where  the  knee  brace  connects  with  the  bottom  chord, 
there  are  now  but  two  unknown  forces  ee^  and  e.K.  From  the  point 
e  in  stress  diagram,  a  line  is  drawn  parallel  with  member  ee^;  and 
from  the  point  K,  a  line  parallel  with  member  e^K,  the  two  lines 
intersecting  in  the  point  elf  ee^  acts  away  from  the  panel  point, 
indicating  tension;  and  e^K  also  acts  away  from  it,  indicating 
tension.  At  panel  point  EF  there  are  three  unknown  forces. 
Ff,  //!  and  /^,  but  the  force  polygon  for  this  joint  may  be 
completed  by  drawing  *,/,  of  such  length  that  the  point  /j  will  be 
half-way  between  the  two  parallel  lines  drawn  from  the  points  F 
and  G.  Then  from  the  point  /,  a  line  is  drawn  parallel  with 
member  ffl  which  intersects  the  line  drawn  from  the  point  F  in  the 
point  /.  Ff  acts  towards  the  panel  point,  indicating  compression ; 
//!  acts  away  from  it,  indicating  tension ;  and  M  acts  towards  it, 
indicating  compression.  The  remainder  of  the  stress  diagram  is 
quite  simple  and  requires  no  further  explanation.  The  point  3 
happens  to  coincide  with  the  point  H,  indicating  that  there  is  no 
stress  in  member  H3. 

The  imaginary  struts  are  now  supposed  to  be  removed,  and  the 
stress  diagram  corrected  accordingly.  The  corrections  ^are  shown 
in  dotted  lines  with  the  points  of  intersection  marked  by  letters  in 
parentheses.  From  the  point  K  in  stress  diagram  a  dotted  line 
is  drawn  parallel  with  the  left-hand  knee  brace,  intersecting  the 
line  db  in  the  point  (b).  At  the  foot  of  knee  brace,  the  force  K(L) 
is  required  to  complete  the  polygon  of  forces.  This  force  is  supplied 


HE 
DIVERSITY 


38  BRIDGE  AND  STRUCTURAL  DESIGN. 

by  the  resistance  of  post  to  bending-.  At  the  top  of  post  the  force  BC 
is  increased  to  (J3)C.  The  difference  (B) B  is  equal  to  the  hori- 
zontal force  at  hinge,  multiplied  by  the  distance  from  hinge  to  foot 
of  knee  brace,  and  divided  by  the  distance  from  foot  of  knee  brace 
to  top  of  post  =  4,610  X  -§•  =  8,300  Ibs.  The  horizontal  force  K(L) 
at  foot  of  knee  brace  is  equal  to  the  horizontal  force  at  hinge,  plus 
the  force  (B)  B  at  top  of  post  =  4,610  +  8,300  =  12,910  Ibs.  At 
the  top  of  right-hand  post,  there  is  also  a  horizontal  force  due  to 
the  moment  of  the  horizontal  force  at  hinge  =  8,300  Ibs.,  and 
represented  by  the  dotted  line- .//(/)  in  stress  diagram.  At  the  foot 
of  knee  brace  the  horizontal  force  to  be  resisted  by  the  bending 
value  of  the  post,  is  the  same  as  for  left-hand  post,  and  is  repre- 
sented in  the  stress  diagram  by  the  line  J(f). 

The  wind  stresses  are  all  figured  on  the  truss  diagram  Fig.  45, 
the  sign  +  indicating  compression,  and  the  sign  — ,  tension.  If  the 
stress  diagram  should  not  close  exactly  at  first,  it  would  be  better 
to  work  from  both  ends  of  truss  towards  the  centre. 

The  maximum  bending  moment  in  the  posts  is  at  the  point  of 
knee-brace  connection,  and  is  equal  to  the  horizontal  reaction  at 
hinge  multiplied  by  its  distance  from  this  point,  =  4,610  Ibs.  X  9  ft. 
=  41,490  ft. -Ibs.  =  497,880  in. -Ibs. 

Stresses  Due  to  Vertical  Loads. — The  total  vertical  load  on  truss 
=  40'  X  16'  X  40  Ibs.  =  25,600  Ibs.  The  intermediate  panel  loads 
=  25,600  -5-  8  =  3,200  Ibs.,  and  the  end  panel  loads  =  i, 600  Ibs. 

Fig.  46  is  a  diagram  of  one-half  of  the  truss,  with  stress  diagram 
for  vertical  loads.  The  stresses  due  to  vertical  loads  which  are 
marked  "V,"  as  well  as  the  maximum  wind  stresses  which  are 
marked  "  W,"  are  shown  on  the  truss  diagram,  and  those  of  the 
same  kind  added  together. 

Bending  in  Rafters. — In  addition  to  direct  compression  in  rafters, 
there  are  bending  moments  due  to  the  loads  which,  in  this  case,  are 
uniformly  distributed,  instead  of  being  concentrated  at  the  panel 
points  by  purlins  as  in  Art.  15.  For  the  bending  moment  in  each 
panel  a  total  load  of  60  Ibs.  per  sq.  ft.  will  be  assumed.  The  hori- 
zontal length,  or  span,  from  one  panel  point  to  another  =  5  ft. 
Load  on  span  —  5'  X  16'  X  60  Ibs.  =  4,800  Ibs.  Bending  moment 

for  simple  span  =  -     — g =  3,000  ft.-lbs.     But  these  spans  are 

continuous,  or  fixed  at  the  ends,  which  reduces  the  bending  moment. 
The  points  of  maximum  moment  are  at  the  ends,  or  panel  points, 


DESIGN  OF  A  KNEE-BRACED  BUILDING.  39 

and  the  moment  at  these  points  is  equal  to  two-thirds  of  the  bend- 
ing1 moment  for  a  span  of  the  same  length  but  not  fixed  at  the  ends, 

—  3,000  X  YZ  =  2,000  ft.-lbs.  =  24,000  in.-lbs. 
Proportioning  of  Members. — For  the  rafters  angles  must  be  selected 

of  such  section  that  the  maximum  stress  per  sq.  in.,  due  to  the 
combination  of  direct  compression  and  bending,  shall  not  exceed 
12,000  Ibs.  2  —  5'  X  3  X  TV  angles  will  be  assumed  with  the 
longer  legs  vertical.  Area  =  4.8  sq.  ins.  R  =  3.78.  Then 

Max.  compression  -f-  area  =  36,900  -f-  4.8    =    7,690 
Max.  bending          -f-  R      =  24,000  -*-  3.78  =    6,350 

14,040  Ibs.  per  sq.  in. 

Since  the  total  fibre  stress  as  above  is  too  great,  2  —  5  X  3  X 
Y%,  Ls  will  be  tried  next.     Area  =  5.72  sq.  ins.     R  =  4.42,  then 

36,900  -*-  5.72  =     6,450 

24,000  -5-  4.42   =      5,430 

1 1, 880  Ibs.  per  sq.  in. 

This  is  satisfactory,  and  the  same  angles  will  be  used  throughout 
the  rafters  to  avoid  splicing. 

In  members  de  and  fg  there  is  a  compression  stress  of  4,650  Ibs., 
and  their  length  is  about  3.3  ft.     Assuming  I  —  2^  X  2  X  ^  L 

1         3,3 
=  1. 06  sq.  ins.  least  r  —  .43".     Then  —  =  ~~  =  7.6,which  by  table 

(Art.  12)  corresponds  to  8,210  Ibs.  per  sq.  in.  for  pin  ends,  and 
4,650  -v-  8,210  =  .56  sq.  ins.  required.  The  area  provided  is  nearly 
double  this  amount. 

In  members  elfl  the  compression  =  13,100  Ibs.,  length  =  6.6  ft. 
2  —  2/^2  X  2>2  X  ^  angles  will  be  assumed,  area  ==  2.38  sq.  ins. 

least  r  —  .77,     —  =  ~  =  8.7,  corresponding  to  7,470  Ibs.  per  sq. 

in.     Then  13,100  -*-  7,470  —  1.75  sq.  ins.  required. 

Members  eel  will  be  made  of  the  same  section  as  ^/,. 

The  knee  braces  must  be  designed  for  —  10,700  Ibs.  or  +  1,6000 
Ibs.     Assuming 2  —  3  X  2,^/2,  X  %  angles  with  the  longer  legs  back 

to  back,  area  =  2.62  sq.  ins.  least  r  —  .95.     /  =  8.3.      —  =  ~^— 

—  8.7,  corresponding  to  a  unit  stress  of  7,470  Ibs.     Then  1,6000  •+• 
7,470  =  2.14  sq.  ins.  required. 


40  BRIDGE  AND  STRUCTURAL  DESIGN. 

The  sections  required  and  those  provided  for  the  tension  members 
are  all  figured  on  diagram  Fig.  46. 

The  posts  must  be  designed  for  bending  stresses  as  well  as  direct 


compression.     A  12  in.  I  @  31.5  Ibs.  is  assumed.     Area  =  9.26  sq, 
ins.     R  =  36,  then 

Direct  compression  -f-  area  =    20,800        Ibs.  -s-    9.26  =    2,250 
Bending  moment     -*-  R      =  497, 880  in. -Ibs. -5- 36       =13,830 

16,080  Ibs.  persq.  in. 


DESIGN  OF  A  KNEE-BRACED  BUILDfNG.  41 

The  maximum  compression  as  above  is  greater  than  that  allowed 
for  the  other  members,  but  since  it  is  nearly  all  due  to  bending  from 
a  maximum  wind  force  which  the  building  will  rarely,  if  ever, 
receive,  and  as  the  posts  are  supported  sidewise  by  the  planking, 
this  unit  stress  is  not  at  all  excessive. 

Anchorage  for  Posts. — The  value  of  one  anchor  bolt  has  been 


taken  at  7,800  Ibs.,  and  it  must  be  let  down  into  the  foundation  far 
enough  to  develop  an  equal  resistance.  100  Ibs.  per  sq.  in.  is  a 
safe  value  for  the  adhesion  of  cement  mortar  to  iron,  and,  as  the 
circumference  of  a  one-inch  bolt  is  about  3  inches,  the  adhesion 
per  lineal  inch  will  be  300  Ibs.  Then  7,800  -*-  300  =  26  inches  = 
length  of  bolt  required  in  foundation.  It  would  be  better  to  extend 
bolts  into  foundation  somewhat  deeper  than  this,  say  36  inches. 


42  BRIDGE  AND  STRUCTURAL  DESIGN. 

The  width  of  foundation  wall  is  assumed  to  be  24  inches,  and  its 
depth  5  ft.  In  constructing-  the  wall,  a  break  about  24  inches  wide 
should  be  made  at  each  post.  The  anchor  bolts  should  then  be  set 
in  their  proper  position  by  means  of  a  template,  and  the  space  filled 
with  Portland  cement  concrete. 

The  anchor  bolt  in  resisting  the  bending-  moment  at  base  of  post 
tends  to  overturn  the  wall  with  a  moment  equal  to  the  value  of  one 
bolt,  multiplied  by  its  distance  from  the  further  edge  of  base  plate 
=  7,800  Ibs.  X  18  ins.  =  140,400  in.-lbs.  This  overturning  mo- 
ment is  resisted,  partly  by  the  weight  of  the  wall,  and  partly  by  the 
earth  filling-  around  it.  The  resistance  of  the  wall  is  easily  figured, 
but  that  of  the  earth  filling-  is  very  indefinite  and  uncertain,  so  the 
latter  will  be  neglected  in  the  present  case.  The  weight  of  wall 
required  is  equal  to  the  overturning  moment  divided  by  the  dis- 
tance from  centre  of  wall  to  its  edge,  =  140,400  in.-lbs.  -=-  12  ins.  = 
11,700  Ibs.  Taking  the  weight  of  masonry  at  150  Ibs.  per  cu.  ft. 
the  weight  of  a  section  of  wall  one  foot  long  =  5'  X  2'  X  150  Ibs. 
=  1,500  Ibs.,  then  11,700  -*-  1,500  =  7.8  ft.,  which  is  the  length  of 
wall  required  to  resist  the  overturning  moment  due  to  anchor  bolts. 
As  the  posts  are  16  ft.  apart,  it  is  evident  that  the  wall  has  ample 
stability  in  itself  without  the  assistance  of  the  earth  filling. 

Fig.  47  is  a  detail  drawing  showing  one-half  of  truss  and  one 
post.  The  general  method  of  designing  details  as  explained  in 
Art.  15  also  applies  to  the  present  example. 

ART.  18.— THE  DESIGN  OF  A  PLATE  GIRDER. 

Data :  Length,  centre  to  centre  of  bearings,  50  ft. 
Depth,  back  to  back  of  angles,  5  ft. 
Load,          4,000  Ibs.  per  lineal  foot. 
Tension,    15,000  Ibs.  per  square  inch. 
Shearing,    7,500  Ibs.  per  square  inch  for   web  plates. 
Shearing,  10,000    "       "         "         "       "    rivets. 
Bearing,  20,000    "       " 
Size  of  rivets  f-in. 

The  bending  moment  at  the  centre  is  given  by  the  formula, 
w  I2  4,000  x  502 

M  = =  =  i, 250,000  ft.-lbs. 

8  8 

The  flange  stress  at  the  centre  equals  the  moment  divided  by  the 
depth,  centre  to  centre,  of  gravity  of  flanges.     When  the  flanges 


THE  DESIGN  OF  A  PLA  TE  GIRDER.  43 

have  cover  plates  the  centre  of  gravity  is  usually  near  the  back  of 
the  angles,  and  sometimes  beyond  that  point,  but  it  is  customary 
to  assume  the  effective  depth  of  such  a  girder  as  the  distance  back 
to  back  of  angles.  As  the  present  example  will  undoubtedly  have 
cover  plates,  the  effective  depth  will  be  5  ft.  Then,  the  flange  stress 
at  centre  =  1,250,000  -f-  5  =  250,000  Ibs.  and  the  net  area  required 
in  bottom  flange  =  250,000  -f-  15,000  =  16.67  square  inches.  When 
practicable,  the  area  of  the  angles  should  be  equal  to  at  least  one- 
third  of  the  total  flange  area. 

The  following  section  will  suit  the  case  : 

Gross  Area.  Eivet  Holes.  Net  Area. 

Two6X3/^X/^  Ls=  9.00         (4X  y%  X  /^  =  1.75) 
Two    isXA-  plates=n.38         (4X^X^=1.52) 

20.38°  17.11 

Allowance  has  been  made  for  two  holes  in  each  angle,  for,  al- 
though the  holes  may  not  come  exactly  opposite  each  other,  their 
stagger  will  not  be  great  enough  to  make  it  allowable  to  deduct  the 
area  of  only  one  hole.  When  rivets  are  staggered  three  inches  or 
more  only  one  hole  need  be  allowed  for. 

Having  decided  on  the  section  of  the  bottom  flange,  it  is  cus- 
tomary to  make  the  top  flange  the  same,  except  in  rare  cases  when 
the  top  flange  is  unsupported  laterally.  It  may  then  be  necessary  to 
make  the  top  flange  wider,  and  to  figure  it  as  a  column.  In  the  pres- 
ent example,  the  top  flange  is  supposed  to  be  supported  laterally  at 
intervals  not  exceeding  fifteen  times  its  width. 

Length  of  cover  plates.  In  Art.  5  it  was  seen  that  the  bending 
moment  at  any  point  of  a  beam  supported  at  both  ends  and  loaded 
uniformly  was  represented  by  ordinates  between  a  parabola,  whose 
vertex  was  at  the  centre,  and  the  closing  line  connecting  the  points 
of  intersection  of  the  parabola  with  verticals  through  the  points  of 
support  of  beam.  Now  the  flange  stress  at  any  point;  also  the  re- 
quired flange  area  may  be  represented  in  the  same  manner. 

Figure  50  represents  one  half  of  the  girder,  which  is  divided  into 
equal  panels  of  5  ft.  In  Fig.  50^  a  parabola  is  constructed  with  a 
base  AB  equal  to  one-half  the  span  of  girder,  and  height  BC  equal 
to  the  area  required  at  centre.  To  construct  the  parabola,  the  line 
AD  —BC  —  16.67  a  "  is  divided  into  the  same  number  of  equal  parts 
as  the  line  AB,  and  from  the  points  of  division  lines  are  drawn 


44 


BRIDGE  AND  STRUCTURAL  DESIGN. 


SQ.Q  cenltc  To  cgntu  of  bearing^ 


Fig.  SI 


4-r 


F/g.  52 


THE  DESIGN  OF  A  PLATE  GIRDER,  45 

to  the  point  C,  as  shown.  The  intersections  of  these  radial  lines  with 
the  verticals  through  the  points  of  division  of  the  line  AB  are  points 
on  the  parabola.  On  this  parabola  are  plotted  the  areas  of  the 
angles  and  cover  plates.  The  angles  extend  the  full  length  of  the 
girder.  The  first  cover  plate  is  required  to  the  point  E,  but  it  is 
customary  to  extend  it  about  one  foot  beyond  this  point,  which 
will  make  the  length  of  the  plate  40  ft.  The  second  cover  plate  is 
required  to  the  point  F.  Adding  one  foot  at  each  end  will  make 
the  length  of  this  plate  28  ft. 

The  following  analytical  method  for  determining  the  proper 
length  of  cover  plates  will  give  the  same  results  as  the  above  graph- 
ical method ;  but  the  graphical  method  is  preferable  on  account  of 
its  simplicity,  and  because  with  it  there  is  much  less  chance  of 
errors. 

In  Fig.  51,  A   =  area  required  at  centre  of  span. 

A!  =  area  required  at  end  of  ist  cover  plate. 

A2  =  area  required  at  end  of  2d  cover  plate. 

X±  =  distance  from  centre'  of  span  to  theoretical  end 

of  ist  cover  plate. 
X2  =  distance  from  centre  of  span  to  theoretical  end 

of  2d  cover  plate. 
1    =  length  of  span,  centre  to  centre  of  bearings. 


ThenX^ 


A  A 

In  the  present  example  A=  16.67  °  ",  At  =7.25  °  ",  A2  =  7.25+4.93 

=  12.18  a",  —  =  25  ft. 


-V 


(16.67 — 7.25)  X252 

-  =  18.8   ft.     Total   length  of    ist 
16.67 

plate  =  (i8.8'X  2)  +  2'  =  39.6  ft.,  say  40  ft. 

(16.67— i2.i8)X  25s 

X2  = • —  =  12.9  ft.  Total  length  of  2nd  plate  = 

16.67 

(12.9X2)  +2'=  27.8  ft.,  say  28  ft. 

The  web  plate  must  have  a  sectional  area  great  enough  to  resist 
the  shear,  and  it  must  be  thick  enough  to  give  sufficient  bearing  for 
the  rivets  in  flange  angles  and  end  stiffeners. 


46  BRIDGE  AND  STRUCTURAL  DESIGN. 

The  maximum  shear  is  at  the  end  and  is  equal  to  one-half  the 

50 
load  on  the  span  =  4,000  Ibs.  x =  100,000  Ibs.    The  permissi- 

2 

ble  shear  per  square  inch  =  7,500  Ibs.  then  100,000  -r-  7,500  = 
13.3  a  "  required.  A  web  plate  60"  X  #"  =  15  °  "  is  all  right  for  the 
shear,  but  may  be  too  thin  for  rivet  bearing,  as  will  be  seen 
presently. 

Rivet  spacing  in  flanges.  The  rivets  connecting  the  web  plate 
with  the  flange  angles  are  required  to  transmit  the  horizontal  shear- 
ing stress  from  the  web  to  the  flanges ;  which  horizontal  shear  in 
any  panel  is  equal  to  the  vertical  shear  at  centre  of  panel  multiplied 
by  its  length  and  divided  by  the  vertical  distance  centre  to  centre  of 
rivets.  When  the  load  rests  directly  on  the  top  or  bottom  flange, 
the  rivets  connecting  this  flange  with  the  web  plate  are  also  re- 
quired to  distribute  the  load.  Then  the  resultant  stress  on  rivets 
of  loaded  flange  is  represented  by  the  hypothenuse  of  a  right 
angled  triangle  in  which  the  other  two  sides  represent  the  horizon- 
tal shear  and  the  vertical  load.  In  the  present  example  the  load  of 
4,000  Ibs.  lineal  foot  is  supposed  to  be  applied  to  the  top  flange. 

Rivet  spacing  in  vertical  legs  of  top  flange  angles  in  panel  ad: 

Vertical   shear   at   centre   of  panel  =  100,000  —  4,000  X  2.5  = 

90,000  Ibs. 
Horizontal  shear  on  rivets  =  90,000  X  -f-f  =  96,400  Ibs.,  then 

96,400-:-  60"  =  1,610  Ibs  per  lineal  inch. 
Vertical  load  on  rivets  =  4,000  Ibs.  per  lineal  foot  =  330  Ibs.  per 

lineal  inch.  

Resultant  stress  on  rivets  per  lineal  inch  =  Vi  ,6io2  +  33O2  =  i  ,640 

Ibs. 
Bearing  value  of  one  Y\  rivet  on  %"  plate  =  3,750  Ibs.     Then 

required  spacing  =  3,750-^-  1,640  =  2.22". 
As  this  is  somewhat  closer  than  desirable,  a  yV  web  plate  will 

be  used  in  this  panel  and  also  in  be. 
Bearing  value  of  one  ^  rivet  on  •&  web  plate  =  4,690  Ibs.    Then 

required  spacing  =  4,690-^  1,640  =  2.86". 
Rivet  spacing  in  vertical  legs  of  top  flange  angles  in  panel  be. 

Vertical  shear  at   centre   of  panel  =  100,000  —  4,000  X  7.5  = 

70,000  Ibs. 
Horizontal  shear  on  rivets  =  70,000  X  •§•§•  =  75,000,  then  75,000 

-s-6o"  =  1,250  Ibs.  per  lineal  inch. 


THE  DESIGN  OF  A  PLATE  GIRDER.  47 


Resultant  stress  on  rivets  per  lineal  inch  =  y  1,250*  +  330*  =  1290 

Ibs. 
Required  spacing-  of  rivets  in  -f^"  web  plate  =  4,690^-  1,290  = 

3.65". 

Rivet  spacing  in  vertical  legs  of  top  flange  angles  in  panel  cd. 

Vertical  shear  at   centre  of  panel  =  100,000  —  4,000  X  12.5  = 

50,000  Ibs. 
Horizontal  shear  on  rivets  =  50,000  X  £g  =  53,600,  then  53,600 

-*•  60"  =  895  Ibs.  per  lineal  inch.  

Resultant  stress  on  rivets  per  lineal  inch =V8952+33O2=  950  Ibs. 
Required  spacing  of  rivets  in  %"  web  plate  =  3,750^-950  =3.94". 
Rivet  spacing1  in  vertical  legs  of  top  flange  angles  in  panel  de. 

Vertical   shear  at  centre  of  panel  =  ioo,ooc —  4,000  X  17.5  = 

30,000  Ibs. 
Horizontal   shear  on  rivets  =  30,000  X  ff  =  32,500  Ibs.,  then 

32,500 -f- 60" =  540  Ibs.  per  lineal  inch. 

Resultant  stress  on  rivets  per  lineal  inch =  V54O2-f-33O2  =  620  Ibs. 
Required  spacing:  of  rivets  in  %"  web  plate =3, 750-^-  620=6.05'  . 

It  is  unnecessary  to  proceed  further  as  the  maximum  spacing 
should  not  exceed  6". 

Rivets  in  flange  plates.  The  ist  flange  plate  requires  a 
sufficient  number  of  rivets  between  its  end  and  the  end  of 
the  2d  flange  plate  to  transmit  to  it  its  full  proportion  of  the 
flange  stress.  The  distance  from  the  end  of  the  ist  plate  to  the 
end  of  the  2d  plate  is  6  ft.  =  72  inches.  The  net  area  of  the  plate 
=  4.93  square  inches ;  then  4.93  x  15,000  =  73,950  Ibs.,  =  its  pro- 
portion of  the  flange  stress.  The  value  of  one  f  rivet  in  single 
shear  =  4,420  Ibs.  Then  73,950  -~-  4,420  =  16  rivets  required  in 
72".  Since  there  are  two  lines  of  rivets  in  the  plate,  the  required 
longitudinal  spacing  =  72" -^  8  =  9".  But  the  pitch  should  not 
exceed  16  times  the  thickness  of  the  plate,  nor  should  it  be  more 
than  6".  The  2d  flange  plate  requires  the  same  number  of  rivets 
between  its  end  and  the  centre  of  the  span  =  14  ft. 

The  rivet  spacing  on  top  and  bottom  flanges  will  be  made  alike 
and  as  uniform  as  possible  in  order  to  simplify  the  template  work. 
In  the  vertical  legs  of  angles  the  spacing  will  be  2/4"  iroma  to  b,  3" 
from  b  to  d  and  6"  from  d  to  /.  In  the  flange  plates  the  spacing  will 
be  6"  throughout,  except  at  splice,  and  the  rivets  will  be  staggered 
with  those  in  the  vertical  legs  of  flange  angles. 


48  BRIDGE  AND  STRUCTURAL  DESIGN. 

Web  splices.  A  60  X  y5^  web  plate  will  be  used  from  a  to  c,  and 
a  60  X  ^  web  plate  from  c  to  /.  The  plates  will  be  spliced  at  c  and  /. 

Web  splice  at  c.  The  shear  at  this  point  =  60,000  Ibs.  The  rivets, 
although  in  double  shear,  have  bearing  on  one  side  of  the  splice  of 
only  J".  The  value  of  one  f  rivet,  bearing  on  J"  plate  =  3,750  Ibs. 
Then,  the  number  of  rivets  required  side  of  splice  adjacent  to  \n 
web  plate  =  60,000  -=-  3,750  =  16.  Two  12"  x  J"  splice  plates  will  be 
used,  and  the  rivets  staggered  as  shown  in  detail,  Fig.  53.  The  same 
number  of  rivets  are  used  on  side  of  splice  adjacent  to  5/16"  web 
plate  for  symmetry  and  to  simplify  the  template  work. 

Web  splice  at  /.  There  is  no  shear  at  this  point.  Two  6"  x  J" 
splice  plates  will  be  used  with  a  single  line  of  rivets  about  6"  apart, 
each  side  of  joint. 

End  Stiffeners.  The  duty  of  the  end  stiffeners  is  to  transfer  the 
shear  from  the  web  plate  to  the  abutments.  They  act  as  columns, 
but  as  the  ratio  of  their  length  to  their  radius  of  gyration  is  small, 
the  unit  stress  of  12,000  Ibs.  per  square  inch  may  be  used  without 
reduction  by  formula.  The  end  shear  or  reaction  =  100,000  Ibs., 
then  100,000  -r-  12,000  =  8.33  square  inches  required  in  end  stiffen- 
ers. Four  5  x  3  x  5/16  Ls  =  9.60  square  inches  will  be  used,  placed 
as  shown  in  Fig.  52,  which  arrangement  is  best  suited  to  distribute 
the  load  over  the  bearing  plate.  The  rivets  are  in  bearing  on  5/16 
plate,  then  the  value  of  one  f  rivet  =  4,690  Ibs.,  and  the  number  of 
rivets  required  =  100,000  -f-  4,690  =  21,  or  n  rivets  in  each  pair 
of  angles.  Between  the  end  stiffeners  and  the  web  plate  3"  x  J" 
fillers  will  be  used  to  avoid  offsetting  the  angles,  and  thus  obtain  a 
better  fit.  These  stiffeners  and  fillers  should  fit  against  the  bottom 
flange  angles  perfectly. 

Intermediate  Stiffeners.  In  case  of  a  heavy  concentrated  load  at 
any  point  of  the  girder,  stiffeners  should  be  proportioned  in  the 
same  manner  as  the  end  stiffeners  to  carry  this  load  and  distribute 
it  into  the  web  plate ;  otherwise,  the  intermediate  stiffeners  are  sim- 
ply to  prevent  the  web  from  buckling,  and  are  usually  spaced  about 
as  far  apart  as  the  depth  of  girder.  There  is  no  scientific  method  of 
proportioning  them,  and  no  generally  accepted  rule.  In  the  pres- 
ent case  two  3  x  3  x  \  Ls  will  be  used.  At  the  top  and  bottom  they 
will  be  offset,  or  crimped,  to  fit  over  the  flange  angles. 

Bearing  Plates.  If  the  girder  rest  on  a  solid  stone  at  each  end, 
the  bearing  pressure  may  be  300  Ibs.  per  square  inch,  but  if  it  be 
supported  on  brickwork,  the  bearing  pressure  should  not  exceed 


THE  DESIGN  OF  A  PLA  TE  GIRDER. 


49 


50  BRIDGE  AND  STRUCTURAL  DESIGN. 

ioo  Ibs.  per  square  inch.  In  the  present  case,  the  former  condition 
will  be  assumed ;  then  the  required  area  of  bearing  plate  =  100,000 
-f-  300  =  333  square  inches.  A  16"  x  f  x  21"  plate  will  be  used  ;  its 
area  =  336  square  inches. 

Flange  Splice.  Angles  and  flange  plates  may  be  obtained  up  to 
sixty  or  seventy  feet  or  even  longer ;  but,  if  the  girder  is  to  be  made 
from  stock  lengths,  it  may  be  necessary  to  splice  the  angles  and  the 
first  cover  plate.  For  the  purpose  of  illustration,  these  will  be  spliced 
near  the  centre  of  the  span  as  shown  in  Fig.  53.  The  net  area  of 
plate  =  4.93  square  inches,  then  4.93  x  15,000  =  73,950  Ibs.  = 
value  of  plate.  The  single  shearing  value  of  one  f  rivet  =  4,420  Ibs. 
then  73,950  -~-  4,420  =  17  rivets  required  on  each  side  of  splice. 
The  drawing  shows  18  rivets.  The  net  area  of  the  angles  =  7.25 
square  inches,  then  7.25  X  15,000  =  108,750  Ibs.  =  value  of  angles, 
and  108,750  -f-  4,420  =  25  rivets  required.  The  drawing  shows 
18  rivets  in  the  6"  legs  of  angles,  and  4  rivets  in  the  3^"  legs,  which 
latter  are  in  double  shear  and  may  be  counted  as  8  rivets.  The 
total  number  of  rivets  in  angle  splice  is  then  18  +  8  =  26.  It 
should  be  noted  that  there  are  about  twice  as  many  rivets 
in  the  6"  legs  as  in  the  3^"  legs.  The  splice  plates  should 
have  at  least  as  much  section  as  the  pieces  spliced.  A  13  x  J  plate 
is  used;  its  net  area,  making  allowance  for  two  J  holes  =  5.62 
square  inches.  In  addition  to  this,  the  vertical  legs  of  the  angles 
are  covered  by  two  3  x  9/16  flats  of  net  area  =  2.39  square  inches. 
Then  5.62  +  2.39  =  8.01  square  inches  in  splice  material  for 
angles,  which  is  somewhat  greater  than  the  net  area  of  the  angles. 
A  common  error  is  to  put  a  long  string  of  rivets  in  a  narrow  splice 
plate,  but  it  will  readily  be  understood  that  it  is  useless  to  use  more 
rivets  in  a  splice  plate  than  are  required  to  develop  its  full  strength. 
Here,  the  two  3  x  9/16  flats  =  2.39  square  inches  net  area, 
and  2.39  x  15,000  Ibs.  =  38,500  Ibs.  In  these  flats  are  4  rivets 
in  full  double  shear  which  are  equivalent  to  8  rivets  in  single  shear, 
then  4,420  x  8  =  35,360  Ibs.  =  value  of  rivets  in  splice  plates. 

ABT.  19.— PLATE   GIBDEB  WITH  ONE-EIGHTH  OF  WEB  PLATE  COMPUTED 

AS  FLANGE  ABEA. 

In  the  foregoing  example,  Art.  18,  it  has  been  assumed  that  the 
bending-  moment  is  resisted  entirely  by  the  flanges,  and  that  the 
web  plate  takes  shear  only.  This  is  not  strictly  correct,  but  is  in 
conformity  with  many  specifications.  It  seems  to  be  better  prac- 


ONE-EIGHTH  OF  WEB  PLATE  AS  FLANGE.  AREA.  51 

tice,  however,  to  make  due  allowance  for  the  resistance  of  the  web 
plate  in  designing  the  flanges. 

The  moment  of  resistance  of  web  plate  =  -g-  =  -7-  h  =  ~T  h. 

In  which  b  =  thickness  of  web  plate. 
h  =  height       "     " 
A"=  area 

Making  allowance  for  a  vertical  line  of  one-inch  holes,  4-inch 

Aw 
centres,  the  net  moment  of  resistance  of  web  plate  =  -g-  h. 

The  moment  of  resistance  of  the  flanges  =  Af  h. 
In  which  A*  =  area  of  one  flange. 

h  =  depth  of  girder  centre  to  centre  of  gravity  of  flanges, 
and  assumed  to  be  equal  to  the  height  of  web 
plate  when  flange  plates  are  used. 
Then,  total  moment  of  resistance  of  girder  = 

(f  h)  +  (A'h)  =  (£  +  A') 


h. 

Therefore,  one-eighth  of  the  area  of  web  plate  may  be  computed 
as  flange  area.  The  following  is  an  alternative  design  for  the 
girder,  using  the  same  shears  and  moments  as  before.  A  -£%"  web 
plate  will  be  used  throughout. 

Flange  area  required  at  centre  =  16.67  sq.  ins. 

Flange  material  provided  = 


yi  of  59#  x  A  web  Plate  =   2-32 

2  —    6  X  Zl/2  X  T7*  Ls  =  7-94  gross  (less  4,  #  holes)  =    6.41 
2  —  13  X    H  plates       =  9.76  gross  (less  4,  #  holes)  =    8.44 

17.17  sq.  ins.  net. 

The  first  cover  plate  requires  to  be  37  ft.  long,  and  the  second 
cover  plate  26  ft.  long. 

Rivet  spacing  in  vertical  legs  of  flange  angles. 

The  longitudinal  shear  per  lineal  inch,  at  any  point  on  the  rivet 
line,  is  equal  to  the  vertical  shear  at  the  point,  divided  by  the 
distance,  in  inches,  centre  to  centre  of  the  rivets  in  the  vertical  legs 
of  the  top  and  bottom  flange  angles  ;  and  the  amount  of  this  shear 
to  be  transferred  by  the  rivets  to  the  flanges  is  proportioned  to 

Area  of  one  flange  at  the  point. 
Area  of  one  flange  +  }i  of  web  plate. 


BRIDGE  AND  STRUCTURAL  DESIGN. 


Rivet  spacing  in  panel  a  b. 

Shear  at  centre  of  panel  =  90,000  Ibs. 

Distance  centre  to  centre  of  rivets  in  vertical  legs  of  top  and 
bottom  flange  angles  =  56  inches. 

Net  area  of  one  flange  at  this  point  =  6.41 
J/b  of  59 >^  X  TV  web  plate  =  2.32 

6.73  sq.  ins. 

Vertical  load  on  rivets  =  330  Ibs.  per  lineal  inch. 
Bearing  value  of  one  ^  rivet  on  -fy  plate  =  4,690  Ibs. 

90,000 


Then,  longitudinal  shear  per  lineal  inch  at  rivet  line    = 

longitudinal  shear  per  lineal  inch  on  rivets  =  1,610  X  ^~J^ 
resultant  stress  on  rivets  per  lineal  inch  = 


=  1,610  Ibs. 

=  i.iSolbs. 
-f-3308  =  1,220  Ibs. 


6.41 


required  spacing  of  rivets  in  top  flange 


inches. 


The  required  rivet  spacing  in  the  remaining  panels  is  found  simi- 
larly. The  required  rivet  spacing  in  flange  plates  is  determined  as 
before. 

Web  Splices. — Since  one-eighth  of  the  web  plate  has  been  com- 
puted as  flange  area,  the  splices  must  be  capable  of  resisting  the 
full  amount  of  bending  moment  attributed  to  the  web,  as  well  as 

the  vertical  shear  at  the  point. 
Bending  value  of  web  plate  = 
2.32  sq.  ins.  X  15,000  Ibs.  X  60 
ins.  =  2,088,000  in.-lbs. 

The  moment  of  resistance  of 
the  splice  must  be  equal  to  or 
greater  than  that  of  the  web. 

Horizontal  splice  plates  8 
inches  wide  will  be  used  adjacent 
to  the  flange  angles,  and  vertical 
splice  plates,  12  inches  wide  be- 
tween them,  as  shown  in  Fig.  54. 

The  number  and  spacing  of  the  rivets  will  first  be  assumed,  and 
then  their  value  investigated. 

The  maximum  •  bearing  value  of  one  ^  rivet  on  -£$  web  =  4,690 
Ibs.,  but  its  value  in  resisting  bending  moment  when  located  in 
neutral  axis  of  girder  is  zero. 


Fig.  64. 


ONE-EIGHTH  OF  WEB  PLATE  AS  FLANGE,  AREA. 


53 


The  distance  from  neutral  axis  to  top  or  bottom  of  girder  =  30 
inches.     Then  the  value  of  one  rivet,  one  inch  from  neutral  axis  = 

A    f~\C\C\ 

'        —  156  Ibs.     The  value  of  any  rivet  will  be  equal  to  156  Ibs., 

multiplied  by  its  distance  from  the  neutral  axis,  and  its  moment  of 
resistance  will  be  equal  to  156  Ibs.,  multiplied  by  the  square  of  this 
distance.  Then  taking-  all  the  rivets  in  splice  plates  on  one  side  of 
the  joint,  both  above  and  below  the  neutral  axis,  their  moment  of 
resistance  is  as  follows  : 


4  rivets  X  156 

Ibs.  X    4tf  ' 

8  =    11,270 

4      "      X    " 

"    X    8>£! 

3  =    45,080 

4      "      X    " 

"    X  i2^! 

J  =  101,430 

4      "      X    " 

"    X  172 

=  180,320 

8  rivets  X  156  Ibs.  X  2O8  =  499,200 
6  "  X  "  "X  22K8  =  473.8oo 
8  "  X  "  "X  258  =  780,000 


338,100  in.  -Ibs.  =  moment  of 
resistance  of  rivets 
in  vertical  plates. 


1,753,000  in.  -Ibs.  =  moment  of 
resistance  of  rivets 
in  horizontal  plates. 

2,091,100  in.  -Ibs.  =  total  mo- 
ment of  resistance 
of  rivets  in  splices. 


The  net  area  of  horizontal  splice  plates  must  be  such  that  their 
moment  of  resistance  shall  be  as  great  as  that  of  the  rivets  in  same, 
viz.,  1,753,000  in.  -Ibs.  The  permissible  tension  for  the  flanges  of 
the  girder  =  15,000  Ibs.  per  sq.  in.  30  inches  from  neutral  axis; 
then,  at  centre  of  horizontal  splice  plates,  22  /^  ins.  from  neutral 

22.5 
axis,  the  permissible  tension  —  15,000  X    —  —  =   11,250  Ibs.   per 

square  inch.     And 

11,250  X  net  area  of  plates  X  22.5  inches  =  1,753,000  in  .-Ibs. 


Therefore,  net  area  of  plates  = 


' 


=  6.9  sq.  ins. 


11,250  /s  22.5 

for  upper  and  lower  plates  together;  or  3.5  sq.  in.  for  one  pair  of 
plates.  Two  8  X  •&  plates  will  be  used.  Their  net  area,  allowing 
for  two  7/&  holes  in  each,  =  3.9  sq.  in. 


54  BRIDGE  AND  STRUCTURAL  DESIGN. 

ART.  20.— DESIGN  FOR  A  WARREN  GIRDER  HIGHWAY  BRIDGE. 
(Figs.  55  and  56.) 

Data:  Length,  centre  to  centre,  of  bearings,  50'  o".  4  panels  of 

12' 6". 

Depth,  centre  to  centre  of  chords,  6'  o". 
Roadway  16'  o"  clear.    Width,  centre  to  centre  of  trusses, 

17'  o". 

Dead  load  (wooden  stringers  and  floor  planking) . . .  250 
Dead  load  (steel) 150 

Total  (pounds  per  lineal  foot) 400 

Live  load,    80  Ibs.  per  square  foot  of  roadway.    Then  80 

Ibs.  x  1 6'  =  1,280  Ibs.  per  lineal  foot. 
Tension,  15,000  Ibs.  per  square  inch. 

Compression,  1,200  Ibs.  per  square  inch,  reduced  by  Ran- 
kine's  formula.  Top  chords  to  be  considered  as  columns 
with  square  ends,  and  web  members  as  columns  with  pin 
ends.    No  compression  member  shall  have  a  length  ex- 
ceeding 120  times  its  least  radius  of  gyration. 
Rivet  shearing,  7,500  Ibs.  per  square  inch. 
Rivet  bearing,  15,000  Ibs.  per  square  inch. 

In  determining  the  dead  load,  as  above,  the  floor  is  supposed  to 
consist  of  3"  plank,  laid- on  3  x  12  joists  about  2'  centres,  and  esti- 
mated to  weigh  3  Ibs.  per  foot  (board  measure).  The  weight  of  steel 
per  lineal  foot  is  given  by  the  formula  2  x  L  +  50  in  which  L  = 
length  of  span  in  feet.  Then  (2  x  50')  +  50  =  150  Ibs.  per  lineal  foot. 

400 

The  dead  load  per  panel  for  one  truss  = Xi2.s'=  2,500  Ibs. 

2 

1,280 

The  live  load  per  panel  for  one  truss  = X  12.5'=  8,000  Ibs. 

2 

These  loads  are  supposed  to  be  concentrated  at  the  lower  panel 
points,  c,  e  and  g. 

The  length  of  the  diagonal  members  =  V6*+6.252  =  8.6/. 

The  stress  in  any  diagonal  is  equal  to  the  shear  in  the  panel  in 
which  it  is  situated,  multiplied  by  the  length  of  diagonal  and  divided 
by  depth  of  truss ;  and  the  shear  in  any  panel  is  equal  to  the  end 
reaction,  minus  any  loads  between  this  end  and  the  panel  considered. 

The  stress  in  any  chord  section  is  equal  to  the  bending-  moment 
at  the  point  where  the  diagonals  in  the  panel  intersect  the  opposite 
chord,  divided  by  the  depth  of  truss. 


WARREN  GIRDER  HIGHWA  Y  BRIDGE.  55 

The  dead  load  stresses  will  be  considered  first. 

The  reaction  at  either  end  is  equal  to  one-half  the  load  on  span, 
or  one  and  one-half  panel  loads, =2, 500 X  1/^  =  3,750  Ibs. 

The  shear  in  panel  ac  is  equal  to  the  reaction  at  a,  and  the  stresses 
in  members  aB  and  Be  which  lie  in  this  panel  will  be  equal  but  of 
opposite  kind.  Thus  aB  will  be  in  compression  and  Be  in  tension. 

The  shear  in  panel  ce  is  equal  to  the  reaction  at  «,  minus  the  load 
at  ct  and  the  stresses  in  cD  and  De  are  also  equal  but  opposite. 

For  the  stress  in  ac  moments  are  taken  about  the  point  B,  which 
is  distant  6.25'  horizontally  from  a.  The  moment  at  this  point  is 
equal  to  the  reaction  at  a  multiplied  by  its  distance  from  B. 

For  the  stress  in  ce  moments  are  taken  about  the  point  D,  which 
is  distant  18.75'  horizontally  from  a.  The  moment  is  equal  to  the 
reaction  at  a,  multiplied  by  its  distance  from  Dt  minus  the  load  at 
c,  multiplied  by  its  horizontal  distance  from  D. 

For  the  stress  in  BD  moments  are  taken  about  the  point  c,  dis- 
tant 12.5'  from  a,  and  the  moment  is  equal  to  the  reaction  at  a, 
multiplied  by  this  distance. 

For  the  stress  in  DF  moments  are  taken  about  the  point  e,  dis- 
tant 25'  from  a.  The  moment  at  this  point  is  equal  to  the  reaction 
at  a  multiplied  by  its  distance  from  e,  less  the  load  at  c  multiplied 
by  its  distance  from  e. 

With  the  above  explanation  the  following:  table  of  stresses  will 
readily  be  understood  : 

DEAD  LOAD  STRESSES. 


Shear  in  panel  0^=3750  —       o  =  3750 

*•  "      ^=3750  —  2500  =  1250 

Moment  at  B       =375°X  6.25'  =23427 
.75'— 


^=3750x12.5'  =46875 

e=3  750X25.0'— 

2500X  I2.5'=62400 


8.67 

Stress   in   aB  Bc=  375OX—  = 

6         5420 
8.67 

"        "   cD  De=  I250X =  1810 

6 

"  "  ac    =23437X  i  =  39io 

"  "  ce     =54687X  "  =  9110 

«  «  BD  =46875X  "  =  7810 

"  «  DF  =624oox  "  =10400 


The  diagonals  in  the  end  panels  as  well  as  the  top  and  bottom 
chords  throughout,  will  receive  their  maximum  live  load  stresses 
when  the  bridge  is  fully  loaded ;  but  the  maximum  stresses  in  the 
intermediate  diagonals  is  caused  by  unsymmetrical  loading;  thus 
cD  will  receive  its  greatest  compression,  and  De  its  greatest  tension 
with  live  loads  at  e  and  g  only ;  while  eF  will  receive  its  greatest 
compression,  and  Fg  its  greatest  tension  with  live  load  at  g  only. 


56  BRIDGE  AND  STRUCTURAL  DESIGN. 

LIVE  LOAD  REACTIONS. 

Reaction  a.     Bridge  fully  loaded     =8000 Xi>£= 12,000  Ibs. 
"  Loads  at  e  and^  only  =8000  X    %=  6,000  Ibs. 

"  Load  at  g-  only  =8oooX    /£=  2,000  Ibs. 


LIVE  LOAD  STRESSES. 


8,ooox  12. 5=200,000 


8.67 


Shear  in  panel  a<r=i2,ooo  =  12,000 

"  "        ce=  6,000  =     6,000 

"  "       eg=  2,000  =     2,000 

Moment  at  j9=i2,oooX  6.25'     =  75,000 

"  Z>=i2,ooox  18. 75'— 

8,ooox6.25'=i75,ooo        "       "  eF Fg=  2,ooox — '=  2,890 
"     tr=i2,oooXi2-5'      =150,000 
"        "    ,=12,000x25.0'-  [    ac     =75JoooX   i  =12,500 


Stress  in  a£  £c=i2,ooox =17,340 

6 

"  cD  De=  6,ooox  — 7=  8,670 
6 


i75,ooox   i  =29,170 


=25,000 
200,000  X    i  =33>33o 


In  Fig.  57,  which  represents  one-half  of  the  span,  the  dead  load 
and  live  load  stresses  are  summarized.  The  compression  of  +  2,890 
Ibs.  in  member  eF,  due  to  live  load  at  g,  is  shown  on  member  De> 
for  this  latter  member  would  receive  the  same  stress  with  live  load  at  c 
only.  Since  the  dead  load  stress  in  De  is  — 1810,  the  resultant  com- 
pression will  be  +  2890 — 1810  =  +  1080  as  shown.  Eight-tenths  of 
this  latter  amount,  which  is  called  a  counter  stress,  is  added  to  the 
maximum  tensile  stress.  In  the  same  manner  the  tension  of  — 2,890 
Ibs.  in  member  Fg  is  shown  on  the  corresponding-  member  cD. 

The  required  area  for  the  tension  members  is  obtained  directly 
by  dividing  the  total  stress  by  the  unit  stress  of  15,000  Ibs.  For 
the  net  area  allowance  has  been  made  for  two  seven-eighths  holes 
in  each  angle. 

The  top  chord  is  supported  horizontally  at  intervals  of  12'  6"  by 
means  of  the  vertical  members  which  are  braced  to  the  floor  beams ; 
and  it  is  supported  vertically  at  intervals  of  6'  3".  Thus  it 
requires  greater  stiffness  horizontally  than  vertically.  Two  4  X  3  X  -ft- 
Ls  will  be  assumed,  with  the  shorter  legs  back  to  back  as  shown. 

1       6.25 

The  least  radius  of  gyration  =.86",  then — = =7.3,  which  (by 

r       .86 

table,  Art.  12)  corresponds  to  10,000  Ibs.  per  square  inch  for  square 
ends.  This  unit  stress  will  apply  to  the  whole  top  chord.  Two 
4X3XT8-  Ls  will  be  used  for  BD  and  two  4X3X  ^  Ls  for  DF. 

\      8.67 

Assuming  the  same  section  for  end  posts  as  top  chords, — = 

r       .86 


WARREN  GIRDER  HIGHWAY  BRIDGE. 


=  10.1,  which  corresponds  to  a  unit  stress  of  6,600  Ibs.  per  square 
inch  for  pin  ends.  Two  4X3X^1?  Ls  will  be  used.  For  member 
cD  two  3X2^XX  Ls  are  assumed,  with  the  3"  legs  back  to  back. 


1       8.67 

—  =  --  =9.1,  then  allowable  unit  stress  =7,300  Ibs.     It  will  be 

r        -95 


Fig.  55 


12-6 


-So'-o  c.1o  c.«1  end  bearings - 


Fig.  51 


16.o 


-t7'.oe.toC.. 


Fig.  5  8 

found  that  the  area  of  two  3X2^  =  ^  Ls  is  considerably  greater 
than  required  for  the  stress,  but  if  smaller  angles  were  used  the 
length  would  exceed  120  times  the  least  radius  of  gyration. 

For  member  De  the  same  section  will  be  used,  as  this  is  also  a 


58  BRIDGE  AND  STRUCTURAL  DESIGN. 

compression  member  under  certain  conditions  of  loading,  as 
explained  on  page  56. 

The  verticals  Cc  Ee  have  no  direct  stress  ;  their  duty  is  to  stiffen  the 

top  chord.     For  these  members  two  2^2X2% X %  Ls  will  be  used. 

Floor  Beams,  Fig.  58.    The  dead   load  consists  of  the   floor, 

which  was  assumed  to  weigh  250  Ibs.  per  linl.  ft.,  plus  the  weight 

of  the  beam  itself.     Then 

Dead  load  =    25oX  i2.5'-|-5oo=  3,625 
Live  load  =1,280X12.5'  =16,000 

Total,  19,625  Ibs. 

The  load  extends  over  sixteen  feet  of  the  floor  beam,  which  is  the 
width  of  roadway ;  but  the  effective  length  for  computing  the 
moment  is  the  distance  centre  to  centre  of  trusses=i7  ft.  The  re- 

19,625 
action  at  each  end  = .     The  moment  at  the  centre  is  equal  to 

2 

the  reaction  multiplied  by  one-half  the  span,  less  the  portion  of  the 
load  on  one  side  of  the  centre,  multiplied  by  the  distance  to  its 
centre  of  gravity. 

^19,625  \      719,625        \     19,625 

Floor    beam    moment=(-^— -  Xg.s')— (-^— ?X4')=-      -X 

2  22 

(8.5 — 4)  =44, 1 50  foot-lbs. 

44,150  foot-lbs.  X  12 =529,800  inch-lbs.  Then  529 ,800-^-1 5, 000= 
35.3  =  R  required. 

A  12"  I  @  31.5  Ibs.  will  be  used.     Its  R=36. 

Laterals.  The  lateral  system,  Fig.  56,  is  a  horizontal  truss  of 
50  ft.  span,  and  17  ft.  deep.  There  are  4  panels  of  12'  6"  each. 
Length  of  diagonals  =  V  12.5*  +  I72  =  21.1'.  The  wind  pressure 
is  taken  at  300  Ibs.  per  lineal  foot  of  bridge,  then  a  panel  load  = 
300  x  12.5  =  3,750  Ibs.  As  in  the  vertical  trusses,  there  is  a  panel 
load  at  each  point  c,  e,  and  g,  and  half  panel  loads  at  a  and  «',  which 
latter  do  not  affect  the  stresses.  The  reactions  as  well  as  the  shear 
in  the  end  panel  =  3,750  x  i£  =  5,620  Ibs.  The  stress  in  the  end 

21. 1 

diagonals  =  5,620  x =  6,790  Ibs.,  and  6,790  -r-  15,000  =  .46 

17 

square  inches  required.  One  2\  x  2  x  J  L  will  be  used.  Its  net  area, 
allowing  for  one  J"  hole  =  .84  square  inches.  The  same  section  will 
also  be  used  for  the  next  diagonal. 

Details,  Fig.  59.    Taking  the  shearing  value  of  rivets  at  7,500  Ibs. 


WARREN  GIRDER  HIGHWAY  BRIDGE. 


59 


60  BRIDGE  AND  STRUCTURAL  DESIGN. 

per  square  inch,  and  the  bearing  value  at  15,000  Ibs.  per  square  inch, 
the  value  of  one  f  "  rivet  in  single  shear  =  3,310  Ibs.,  bearing  on  f" 
plate  =  4,220  Ibs.  and  bearing  on  J"  plate  =  2,810  Ibs.  The  num- 
ber of  rivets  required  are  clearly  shown  on  the  drawing,  and  it  is 
only  necessary  to  explain  one  or  two  points. 

The  bearing-  plate  at  a  requires  to  be  large  enough  so  that  the 
pressure  on  the  masonry  shall  not  exceed  300  Ibs.  per  square  inch. 
The  dead  load  reaction  =  3,750  Ibs.  and  the  live  load  reaction 
=  12,000  Ibs.,  making  a  total  of  15,750  Ibs.  Then,  15,750  -=-  300  = 
50  square  inches  required.  The  plate  used,  which  is  12"  x  i6J"  = 
198  square  inches,  is  much  larger  than  necessary  for  bearing  on 
the  masonry.  It  also  requires  to  be  large  enough  for  the  anchor 
bolt  and  to  make  connections  with  the  bottom  chord  angle  and  the 
laterals.  At  B  the  hip  cover  plate  is  added  to  give  more  lateral 
stiffness  at  this  point.  The  bottom  chord  splice  at  c  is  formed  partly 
with  the  gusset  plate,  and  partly  with  the  plate  on  the  bottom.  The 
net  area  through  the  splice  should  not  be  less  than  that  required 
in  the  member  a  c,  viz.,  1.09  square  inches.  It  is  evident  that  the 
whole  width  of  the  gusset  plate  cannot  be  relied  on,  as  the  pull  is  all 
on  one  edge,  and  if  the  plate  were  to  begin  to  fail  at  the  edge,  a 
piece  of  it  would  soon  be  torn  off.  It  is  only  safe  to  figure  on  a 
width  of  plate  equal  to  twice  the  rivet  gauge  in  the  angle,  =  2j", 
and  from  this  width  should  be  deducted  the  diameter  of  the  rivet 
hole,  J".  Then  the  net  area  =  (2.75"  —  .875')  x  f "  =  .70  square 
inches.  The  net  area  of  the  bottom  plate  making  allowance  for 
two  7/s"  holes  =  (7.5  ~  1-75)  X  #  =  1.44  square  inches.  Then 
the  net  area  through  splice  =  .70  +  1.44  =  2.14  square  inches, 
which  is  considerably  greater  than  required.  The  value  of  the  rivets 
at  left  end  of  splice  should  be  at  least  equal  to  the  stress  in 
a  c  =  16,410  Ibs.  Then 

2  rivets  bearing  on  f"  plate  at  4,220  Ibs.  =   8,440 
4  rivets  bearing  on  J"  plate  at  2,810  Ibs.  =  11,240 

Total i9>68o  Ibs. 

The  value  of  the  rivets  at  right  end  of  splice  should  be  equal  to  or 
greater  than  the  stress  c  e  =  38,280  Ibs.  Then, 

7  rivets  bearing  on  f"  plate  at  4,220  Ibs.  =  29,540 
4  rivets  bearing  on  J"  plate  at  2,810  Ibs.  =  11,240 

Total  .        40,780  Ibs. 


WARREN  GIRDER  HIGHWA  Y  BRIDGE.  6 1 

The  splice  in  top  chord  at  D  is  designed  similarly. 

Floor  beam  connection,  Fig.  59a.  The  reaction  or  end  shear  is 
equal  to  one-half  the  total  load  on  floor  beam  =  9,810  Ibs.  The 
rivets  connecting  the  end  angles  with  the  truss  are  in  direct  single 
shear,  therefore  9,810  -f-  3,310  =  3  rivets  required,  whereas  there 
are  4  rivets  provided.  In  addition  to  the  vertical  load  on  the  rivets 
connecting  the  end  angles  with  the  web  of  beam,  these  rivets  are  re- 
quired to  resist  a  bending  moment  equal  to  the  reaction  multiplied 
by  the  distance  from  back  of  angles  to  the  centre  of  gravity  of 
rivets.  The  greatest  stress,  due  to  bending,  is  on  the  rivets  far- 
thest from  their  centre  of  gravity,  and  is  equal  to  the  bending  mo- 
ment, divided  by  the  moment  of  resistance  of  rivets.  The  direction 
of  this  stress  is  perpendicular  to  a  line  drawn  through  the  centre 
of  outer  rivet,  and  the  centre  of  gravity  of  the  system.  The  centre 
of  gravity  of  rivets  and  their  distances  from  this  point  are  shown  in 
Fig.  59a. 

The  bending  moment  =  9,810  Ibs.  x  2.55"  =  25,000  inch-lbs. 

The  polar  moment  of  inertia  is  obtained  by  multiplying  each  rivet 
by  the  square  of  its  distance  from  the  centre  of  gravity  as  follows : 

I  =  i  rivet  x  .8"2  ==  .6 
2  rivets  x  1.922  =  7.4 
2  rivets  x  3.n2  =  19.4 


27.4 

The  moment  of  resistance  is  obtained  by  dividing  the  above  re- 
sult by  the  distance  from  the  centre  of  gravity  to  the  farthest 
rivet,  thus : 

R=  1  =  ^=8.8. 
n        3.11 

Then  the  stress  on  the  outer  rivets  from  bending  =  25,000  inch- 
lbs.  -5-  8.8  =  2,840  Ibs.  The  vertical  load  on  each  rivet  is  equal  to  the 
reaction  divided  by  the  total  number  of  rivets  =  9,810  Ibs.  -f-  5  = 
1,960  Ibs.  The  resultant  stress  =  3,900  Ibs.  and  is  obtained  graph- 
ically as  shown.  It  is  less  than  the  bearing  value  of  a  f "  rivet  on  the 
web  of  beam  which  is  f "  thick,  and  consequently  the  connection  is 
satisfactory. 

Camber.  Bridge  trusses  are  constructed  with  a  slight  arch  called 
camber.  This  adds  nothing  to  their  strength,  and  is  intended  prin- 
cipally to  offset  the  deflection  due  to  the  dead  and  live  loads.  The 


62  BRIDGE  AND  STRUCTURAL  DESIGN. 

camber  is  obtained  by  making  the  top  chord  slightly  longer  than  the 
bottom,  and  increasing  the  length  of  the  diagonal  members  in  pro- 
portion. The  following  rule  is  taken  from  Trautwine : 

8  d  c 


In  which  i  =  total  increased  length  of  top  chord. 
d  =  depth  of  truss. 
c  =  camber  at  centre, 
s  =  span. 

All  in  feet  or  all  in  inches. 
In  the  present  example  a  camber  of  i"  is  assumed,  d  =  6'  =  72", 

8  x  72  x  i 

s  =  50'  =  600",  then  i  = —  .96",  say  i".  Since  there 

600 

are  four  panels  in  bridge,  each  top  chord  panel  must  be  increased 
i",  or  each  half  panel  J".  The  lengths  of  BC,  CD,  DE  will  then  be 
6'  3"  -f-  >6".  To  obtain  the  lengths  of  diagonals,  the  mean 
between  the  top  and  bottom  half-panel  lengths  is  combined  with 
depth  of  truss  thus:  length  of  diagonals  =  V(6/3IV//)2+(6V/)2  = 
8'  8". 


ART.    21.— DESIGN    FOR    SKEW    WARREN    GIRDER    HIGHWAY    BRIDGE. 

(Fig.   60.) 

Data :  Length,  centre  to  centre  of  bearings,  72'  o".    4  panels  of 

15' o"  and  i  panel  of  12' o". 
Depth,  centre  to  centre  of  chords,  7'  6". 
Roadway  i6'o"  clear.    Trusses,  17' 6"  centre  to  centre. 
Dead  load  (wooden  stringers  and  floor  planking) . . .  250 
(steel  =  2  L  +  50  —  (2  X  72')  +  50  =  104)  say  200 

Total  ( pounds  per  lineal  foot ) 450 

Live  load  for  trusses,  75  Ibs.  per  square  foot  of  roadway 
=  1,200  Ibs.  per  lineal  foot. 

Live  load  for  floor  beams  100  Ibs.  per  square  foot  of  road- 
way. 

Horizontal  wind  force,  300  Ibs.  per  lineal  foot,  one-half  of 
which  to  be  treated  as  live  load. 


SKEW  WARREN  GIRDER.  63 

Unit  Stresses:  Tension,  15,000  Ibs.  per  square  inch. 

Compression,  12,000  Ibs.  per  square  inch, 
reduced  by  Rankine's  formula  (Art.  12). 
Top  chords  to  be  considered  as  columns 
with  square  ends,  and  web  members  as 
columns  with  pin  ends.  No  compression 
member  shall  have  a  length  exceeding 
120  times  its  least  radius  of  gyration. 
Rivet  shearing,  7,500  Ibs.  per  square  inch. 
Rivet  bearing,  15,000  Ibs.  per  square  inch. 

PANEL  LOADS  FOB  ONE  TRUSS. 

450 
For  panel  points  c,  e  and  g  dead  load  =  ---  X     15     =3375  Ibs. 

1  200 
"         "    live  load   =  -  X     15     =9000  Ibs. 

2 

•«  "       i  dead  load  =-5°  X—  —=3000  Ibs. 


live  load   =-X  =8100  Ibs. 


2 


Length  of  regular  diagonal  members  =  V7-52  +  7-52  =  10.61'. 
Length  of  diagonals  is  e'^and  Jk  =  V7-52  +  62  —  9.60'. 
The  dead  load  reaction  at  a  is  equal  to  the  moments  of  the  panel 
loads  about  k,  and  divided  by  the  length  of  span  ok. 

\  (3,000  X  12)  }   +  {3,375  X  (27  +  42  +  57)  }  _  , 

—    _  -  —  __  -  -  .  --  0,405   IDS. 

72 

The  dead  reaction  at  k  is  equal  to  the  sum  of  the  panel  loads,  less 
the  reaction  at  a  =  3,000  +  (3,375  x  3)  —  6,405  =  6,720  Ibs. 

In  the  following  table  of  dead  load  stresses,  the  same  general 
method  employed  in  Art.  20  is  observed,  but  both  ends  of  the  truss 
are  considered. 

The  shear  in  panel  ac  is  equal  to  the  reaction  at  a. 

"  "       ce  "  "  "     a,  minus  the  panel  load  at  c. 

eg.  „  «  <.      k         «  «  «< 


The  bending  moments  at  B,  D,  F,  c  and  e  are  obtained  by  taking 


64 


BRIDGE  AND  STRUCTURAL  DESIGN. 


moments  of  the  reaction  a  about  these  points,  and  deducting  the 
moments  of  the  panel  loads  between  these  points  and  a.  For  the 
bending-  moments  at  H,  J,  g  and  i  moments  of  the  reaction  k  are 
taken  about  these  points,  and  the  moments  of  the  panel  loads  be- 
tween these  points  and  k  are  deducted. 


DEAD  LOAD  STRESSES. 


Shear  in  #£=6,405 —       o         =     6,405 
"       "    ^=6,405 — 3,375         =    3,030 
«  ^=6,720— (3,375 

+3,000)=       345 

«       <«    £-7=6,720—3,000        =    3,720 
"       "    zVfe=6, 720- —       o        ==    6,720 
Moment  at  ^=6, 405  X     7-5         =  48,000 
".0=6,405x22.5— 

3,375XV-5=n8,8oo 

"^=6,405x37-5- 

3>375X(7.5+22.5)=i38,95o 

"^=6,720X19.5— 

3,000X7.5=108,550 
"/=6,720X  6  =  40,300 

"^=6,405x15  =96,000 

"        "  e  =6, 405X30— 

3,375Xi5=i4i,5oo 


3,000x15= 

12  =  80,650 


Stress  in  aB  Bc=  6,4O5X 

1U.O1 

T7 

=  9,000 

"     cD  De=    3,030X     ** 

=  3,200 

«« 

*FFff=       345  X 

K 

=    500 

" 

gH  Hi 

=    3,72ox 

" 

=  5,250 

« 

UJk 

=    6,72ox 

9.6O 

=  8,600 

7-5 

i 

«< 

ac 

=  48,ooox 

=  6,400 

7-5 

« 

ce 

=n8,8oox 

« 

=15,800 

«« 

eg 

=i38,95°X 

«« 

=18,500 

ii 

Si 

=io8,55ox 

« 

=14,400 

" 

ik 

=  4o,3oox 

tt 

==  5,400 

« 

BD 

=  96,ooox 

<« 

=12,800 

" 

DF 

=i4i,5oox 

ii 

=18,800 

" 

FIT 

=i36,45ox 

<« 

=18,200 

« 

HJ 

=  8o,65ox 

«  < 

=10,700 

The  maximum  live  load  stresses  in  top  and  bottom  chords,  and  in 
diagonals  aB,  Be,  iJ  and  Jk  will  occur  when  the  bridge  is  fully 
loaded.  The  maximum  compression  in  cD  and  tension  in  De,  with 
loads  at  e,  g  and  /  only.  The  maximum  compression  in  eF  and 
tension  in  Fg  with  loads  at  g  and  i  only.  The  maximum 
tension  in  eF  and  compression  Fg  with  loads  at  c  and  e 
only.  The  maximum  tension  in  gH  and  compression  in  Hi  with 
loads  at  c,  e  and  g  only.  In  computing  the  live  load  reaction  a,  mo- 
ments are  taken  about  k;  and  in  computing  the  reactions  k,  mo- 
ments are  taken  about  a,  except  for  case  of  bridge  fully  loaded, 
when  reaction  k  is  equal  to  the  total  load  on  span,  less  the 
reaction  a. 


Reaction  a.     Bridge  fully  loaded  = 


SKEW  WARREN  GIRDER. 
LIVE  LOAD  REACTIONS. 

[SiooX  12]  4-  [90ooX(27+42+57)l 


72 


[8100X12]  +  [ooooX(27+42)] 
a.     Loads  at  e,  g  and  t= 


a.     Loads  at  g-  and  t      = 


72 

(8100X12)4-  (9000X27) 
72 


k.     Bridge  fully  loaded  =8100+  (9000X3)— 17100 

9oooX(i5+3o+45) 
k.     Loads  at  c,  e  and  g  =  — 


:.     Loads  at  c  and  e      = 


72 

9ooox(i5+3o) 
72 


65 

=17,100 
=  9,970 

=  4,725 
=18,000 

=11,250 
=  5,625 


LIVE  LOAD  STRESSES. 


Shear   in  ac  =17,100  *'• 

"       "     ce=  9,970 
«       «     <gr  =  4,725  ?*. 

«         «      ^          5>625 

"         "     ^'=11,250 
«         "      /ft  =18,000 

Moment  at  ^=17,  iooX   7-5 
»        "  Z>=i7,  100x22.  5— 

9,oooX7-5= 


9,ooox(7-5+22.5> 
"^7=18,000x19.5— 

8,iooX7-5= 
"/=i8,ooox  6 
"  ^=17,100X15  = 

41  ^=17,100x30— 

9,000x15= 
"  g  =18,000x27— 

8,100x15= 
«'  z  =18,000X12        ,».« 


17,100 
9,970 
4,725 
5,625 
11,250 
18,000 
128,000 

317,500 
371,250 
290,250 

108,000 

256,500 
=378,000 
=364,500 

=216,000 

10.  61 
Stress  \naBBc=  1  7,  iooX  =24,  100 

/  '  *) 

"      cDDe=    9,970X     "    =14,050 
««     eFFg=    4,725X     "    ==  6,650 
"     eFFg=    5,625X     "    ==  7,95° 
"    gHHi=  ii,2sox     "    =15,850 
9.60 

«         •/  /^  /"^  —  ^    rR  rw*»\/                —  ^*5  r^Cri 

V  y*  —   io,uoo  A            —  ^jjUyj 

oc    —  i28,ooox          —  17,05^ 

ce     =3i7,5oox     "    =42,300 
*ff    =37i,25QX     "    =49,400 
gi    =290,250X     "    =38,700 
"        ik    =io8,ooox     "    =14,400 
"       BD  =256,5oox    "    =34,200 
DF  =378,ooox     "    =50,400 
"       FH  =364,5ooX     "    =48,500 
HJ   =2i6,oooX     "    =28,800 

Since  the  truss  is  unsymmetrical,  it  is  necessary  to  make  a  com- 
plete stress  diagram,  Fig.  61.  The  various  members  are  propor- 
tioned in  the  same  manner  as  in  previous  example  of  5O-ft.  span. 
As  the  present  span  is  much  longer  than  the  last,  the  vertical  legs 
of  angles  are  turned  out  as  shown  to  give  greater  stiffness  horizon- 


66 


BRIDGE  AND  STRUCTURAL  DESIGN. 

I/ 


fc 

A 

<»• 

V9 

H 

^» 

4i 

€ 

11 

I 

X 

r 

<   c 

•R 

-0- 

3 

IM* 

i  ' 

M 

J  4 


SKEW  WARREN  GIRDER. 


tally.    This  will  necessitate  double  gusset  plates,  and  all  angles  will 
require  to  be  latticed. 

Floor  Beams,  Fig.  62: 

Dead  load  (floor)  =  250  Ibs.  x  15'  =  3,750 
Dead  load  (beam)   =  say  750 


Live  load  =  16'  x  15'  x  100  Ibs.  = 


Total 


Reaction  = 


4,500 
24,000 

28,500  Ibs. 


28,500 


=  14,250  Ibs.     Moments  at  centre   = 


(14,250  x  8.75)  —  (14,250  x  4)  =  14,250  x  (8.74  —  4)  =  67,700  ft.-lbs. 
67,700  x  12  =  812,400  inch-lbs.  Then,  812,400  -=-  15,000  =  54.16  = 
R  required. 

A  15"  I  at  42  Ibs.  will  be  used.    R  =  58.9. 

Laterals,  Fig.  60.  It  will  be  sufficiently  accurate  to  assume  that 
the  horizontal  truss  consists  of  five  equal  panels  of  1 5'  o"  each,  and 
17'  6"  deep.  The  length  of  the  diagonals  =  VJ52  +  J7'  6"2  =  23.04 
ft.  One-half  the  wind  force  is  to  be  treated  as  a  stationary  or  dead 
load,  and  one-half  as  a  moving  or  live  load,  then 

Panel  dead  load  =150  Ibs.  x  15'  =  2,250  Ibs. 
"       live      "    =  150  Ibs.  x  15'  =  2,250  Ibs. 

The  maximum  shear  in  panel  a  c  is  when  the  live  load  covers  the 
span. 

The  maximum  shear  in  panel  c  e  is  when  the  live  load  is  at  e,  g 
and  i  only. 

The  maximum  shear  in  panel  e  g  is  when  the  live  load  is  at  g  and 
i  only. 


ist   lateral   = 


DEAD  LOAD  STRESSES. 
23.04 


5,94O 


2d         "       =   2,75oXiX     "       =   2,970 
3rd       "       =  ==          o 


2,250  X  2  X 

2,250  X  |  X 
2,250  X  |  X 


LIVE  LOAD  STRESSES. 
23.04 


17-5 


=  5,940 

=  3,560 
=  1,780 


The  total  stresses,  the  areas  required  and  provided  are  shown  in 
Fig.  60. 

Floor  beam  connection,  Fig.  63.  The  reaction  =  14,250  Ibs.  and 
the  rivets  connecting  the  end  angles  with  the  truss  are  in  direct 


68  BRIDGE  AND  STRUCTURAL  DESIGN. 

shear;  therefore,  14,250  -5-  3,310  =  5  rivets  required,  whereas  the 
drawing  shows  6  rivets.  In  addition  to  the  vertical  load  on  rivets 
connecting  the  end  angles  with  web  of  beam,  these  rivets  are  re- 
quired to  resist  a  bending  moment  equal  to  the  reaction  mulitplied 
by  the  distance  from  back  of  angle  to  centre  line  of  rivets.  The 
greatest  stress,  due  to  bending,  is  on  the  rivets  farthest  from  their 
centre  of  gravity  and  is  equal  to  the  bending  moment,  divided  by 
the  moment  of  resistance  of  rivets.  The  direction  of  this  stress  is 
horizontal.  The  centre  of  gravity  of  rivets  and  their  distances  from 
this  point  are  shown  in  Fig.  63. 

The  bending:  moment  =  14,250  Ibs.  X  1.75"  =  24,900  in.-lbs. 

The  moment  of  inertia  of  rivets  is  obtained  by  multiplying  each 
rivet  by  the  square  of  its  distance  from  the  center  of  gravity,  as 
follows : 

1  =  2  rivets  x  1.52  =    4.5 
2  rivets  x  4.52  =  40.5 

45-0 

The  moment  of  resistance  is  equal  to  the  above  result  divided 
by  the  distance  from  centre  of  gravity  to  farthest  rivet,  thus : 

45 

4-5 

Stress  on  outer  rivets  from  bending  =  24,900  in.-lbs.  -f-  10  = 
2,490  Ibs. 

Stress  on  outer  rivets  from  direct  load  =  14,250  Ibs.  ~-  4  = 
3,560  Ibs.  ^ 

Resultant  stress  on  outer  rivets  =  V  2,49O2  +  3,5602  =  4,340  Ibs. 

The  web  of  beam  is  £f  "  thick,  therefore,  the  bearing  value  of  one 
|  rivet  =  £f"  x  f  x  15,000  Ibs.  =  4,570,  which  is  greater  than  the 
maximum  stress  on  rivets. 

The  other  details,  Fig.  64,  are  self-explanatory. 

ART.  22.— DESIGN  FOR  A  PIN-CONNECTED  PRATT  TRUSS  HIGH- 
WAY SPAN.      (FIG.   65.) 

Data:   Length,  120'  o",  centre  to  centre  of  end  pins.    8  panels 

of  15'  o". 
Depth,  20'  o". 
Roadway,  16'  o"  clear.    Trusses,  17'  o",  centre  to  centre. 


SKEW  WARREN  GIRDER. 


* .* 


70  BRIDGE  AND  STRUCTURAL  DESIGN. 

Dead  load  (steel)  =  2  x  L  +  50  — 290 

"     (floor)  =  250 

Total,  per  lineal  foot  = 540  Ibs. 

Live  load,    75  Ibs.  per  square  foot.    Then  75  Ibs.  x  16'  = 

1,200  Ibs.  per  lineal  foot  for  trusses. 
Live  load,  100  Ibs.  per  square  foot  for  floor  beams  and 

hip  verticals. 
Wind  force  for  top  laterals,  150  Ibs.  per  lineal  foot,  to  be 

treated  as  dead  load. 
Wind  force  for  bottom  laterals,  150  Ibs.  per  lineal  foot,  to 

be  treated  as  dead  load. 
Wind  force  for  bottom  laterals,  150  Ibs.  per  lineal  foot  to 

be  treated  as  live  load. 

STEEL  IRON 

Unit  stresses:  Tension. 1 5,000     12,000  for    bottom    chords,    main 

diagonals       and       floor 
beams. 

9,000  for  counters  and  hip  ver- 
ticals.. 

8,000  for  floor  beam  hangers. 
15,000  for  laterals. 

STEEL 

Compression,  12,000  reduced  by  Rankine's  formula  (Art.  12).  No 
member  to  have  a  length  exceding  120  times 
its  least  radius  of  gyration ;  end  and  inter- 
mediate posts  to  be  considered  as  columns 
with  two  pin  ends ;  end  sections  of  top  chord 
as  columns  With  one  pin  and  one  square 
end;  intermediate  sections  of  top  chord  as 
columns  with  square  ends. 

Shearing,         10,000  for  pins  and  shop  rivets. 
"  7>5OO  for  field  rivets. 

Bearing,  20,000  for  pins  and  shop  rivets. 

"  15,000  for  field  rivets. 

Bending,          22,500  for  pins. 

PANEL  LOADS  FOB  ONE  TRUSS. 


540 


Dead  Load  = Xi5'  =  4,000  Ibs.         Length  of  diagonals  =  -v/i58-h2o8  =  25'. 

1 200 
Live  Load   = Xi5'  =  9,000  Ibs. 

Dead  Load  Reaction  =  4000X3^  =  14,000  Ibs. 


PIN-CONNECTED  HIGHWA  Y  SPAN.  ' 


DEAD  LOAD  STRESSES. 


Shear  in  panel  ab=  14,  ooo 


Moment  at    B 


=14,000 
4,000=10,000 
^£=14,000  —  8,000=  6,000 

cfe=I4,000  —  12,000=    2,000 

"=  1  4,000—16,000=—  2,OOO 

=i4,oooXI5      =210,000 

14,000X30— 

4,oooX  15=360,000 
=  14,  oooX4S  — 
4,ooox(i5+3o)=45o,ooo 
e    =14,000X60  — 
4,ooox(i5+3o+45)=48o,ooo 


Stress  in  aB 
Be 
Cc 
Cd 
Dd 
De 
Ee 
Ef 
abc 


17,500 
io,oooxf£=  12,500 
6,oooX  i  =    6,000 
6,oooxf£=     7>5oo 

2,OOOX  I  =  2,OOO 
2,OOOXf<T=  2,500 
2,OOOX  I  =  —  2,000 


DE     = 


210,000X^7=    10,500 

3  60,  ooo  X  A—  18,000 
=  22,500 
=  24,000 


For  the  stress  in  a  b  c  moments  are  taken  about  panel  point  B. 
For  the  stress  in  B  C  moments  are  taken  about  c,  and  for  the  stress 
c  d  moments  are  taken  about  C.  But  since  c  and  C  are  the  same 
distance  horizontally  from  a,  the  stresses  in  B  C  and  c  d  are  equal. 
Likewise  the  stresses  C  D  and  d  e  are  equal. 


LIVE  LOAD  REACTIONS. 

Bridge  fully  loaded  9,oooX3K==3I»5°o  Iks.  for  maximum  stresses  in  chords  and 

end  posts. 

Loads  at  c,  d,  e,f,  g  and  7i,  9,oooX\1-==23,6oo  Ibs.  for  maximum  stress  in  Be. 
Loads  at  d,  e,f,  g  and  h,  g,oooy^i/-=i6,goo  Ibs.  for  maximum  stresses  in  Cc  and  Cd. 
Loads  at  e,f,  g  and  h,  g,ooo'X™=ii,2$o  Ibs.  for  maximum  stresses  in  Dd  and  De. 
Loads  at/,  g  and  h,  9,000x1=6,750  Ibs.  for  maximum  stresses  in  Ee  and  Ef. 


LIVE  LOAD  STRESSES. 


Shear  in  panel  ab 

"  "      be 

"      cd 

"  "      de 

"       ef 

Moment  at    B    31,500X15 
"         "  ^C  3  1,  500x30— 
9,000x15 


9,ooox(i5+3o) 
"  e  31,500X60— 
9,ooox(i5+3o+45) 


31,500 

Stress  in 

a^  = 

3i,5ooxf£=39,400 

23,600 

« 

Be    = 

23,600X^=29,500 

16,900 

« 

Cfc    = 

i6,90QX  I  =16,900 

11,250 

« 

Cd   = 

16,900x1^=21,100 

6,750 

« 

Dd  = 

n,25ox  i  =11,250 

472,500 

« 

/?<?    = 

11,250X^=14,050 

" 

Ee    = 

6,75°X  i  =  6,750 

810,000 

« 

^/   = 

6,75oXf&=  8,450 

« 

abc   = 

472,5oox  20  =23,625 

1,010,000 

M 

BCcd= 

8  10,  ooo  X  "=40,500 

<« 

CD  d£=i 

,oio,oooX  "=50,500 

1,075,000 

« 

DE   =i 

,o75,ooox  "=53,75o 

Hip  Vertical  Bb.    The  live  load  for  the  hip  vertical  is  taken  at 


72  BRIDGE  AND  STRUCTURAL  DESIGN. 

100  Ibs.  per  square  foot  of  roadway  =  1,600  Ibs.  per  lineal  foot  of 
bridge.     The   stress   in   this   member  will  then  be  -  -  X  15'  = 

12,000  Ibs. 

The  stresses  and  material  are  shown  in  Fig.  65.  Square  iron  bars 
are  used  for  all  tension  members  requiring  less  than  four  square 
inches,  and  flat  steel  bars  for  all  others.  The  square  bars  have  loop 
ends  which  must  be  welded,  and  steel  is  not  suitable  on  this  account, 
as  it  is  difficult  to  weld  it  satisfactorily.  The  heads  on  the  flat  steel 
bars  are  upset  and  forged  or  pressed  into  dies  without  welding. 
The  intermediate  posts  have  a  much  greater  area  than  is  required 
for  the  stress,  but  with  smaller  channels  the  length  would  exceed 
1  20  times  the  least  radius  of  gyration,  and  the  lighter  weight  of  the 
6"  channels  has  a  web  of  only  .20  inch,  which  is  entirely  too  thin. 

The  area  of  top  chords  is  also  rather  large,  but  smaller  channels 
are  unadvisable.  The  end  posts  will  be  considered  in  connection 
with  portal  bracing. 

Top  Laterals,  Fig.  65.  The  top  lateral  truss  consists  of  six  15-ft. 
panels,  17.25  ft.  deep.  Diagonal  =  V  J52  +  I7-252  =  22-8  ft-  Panel 
.  =  150  Ibs.  x  15  ft.  =  2,250  Ibs. 

STRESSES. 

22.8 

ist  lateral=225ox  2>£  X  -  =75oo  Ibs. 

22.8 

2nd     "    =22oXiX---=45oo  Ibs. 


22.8 
8rd      "     =2250  X  #  X  -  =1500  Ibs. 

Bottom  Laterals.  The  bottom  lateral  truss  consists  of  eight 
15  ft.  panels  17.25  ft.  deep.  Diagonal  =  V  IS2  +  i7-252  —  22.8  ft. 
Dead  panel  load  =  live  panel  load  =  150  Ibs.  X  15  ft.  =  2250  Ibs. 


DEAD  LOAD  STRESSES. 
22.8 

ist  lateral  =  2,2$oX3#X  -  =  10,500 

I7-25 
2nd      "     =  2,250x2^  X     "     =    7,5oo 


2,250x3^  Xi7~io,  500 

2,25oX  V-  X     "    =  7,900 
2,25ox  V-  X     "=  5,600 

2,250X  Y  X      "     =3,700 

Portal  Strut  and  End  Posts,  Fig.  65.    It  is  assumed  that  the  wind 
force,  which  is  equal  to  three  and  one-half  top  lateral  panel  loads,  is 


3rd      «'     =  2,25oxi^X     "     =    4,500 
4th       "     =  2,25ox  >£X     "     =    1,500 


LIVE  LOAD  STRESSES. 


PIN-CONNECTED  HIGHWA  Y  SPAN. 


73 


74  BRIDGE  AND  STRUCTURAL  DESIGN. 

applied  at  the  top  of  portal  strut,  and  that  it  is  resisted  equally  at  the 
foot  of  both  end  posts.  It  is  also  assumed  that  the  posts,  although 
hinged  at  the  bottom  in  plane  of  trusses,  are  fixed  in  the  plane  of 
the  portal,  and  that  the  plane  of  contra-flexure  is  midway  between 
the  foot  of  posts  and  the  lower  extremities  of  portal  struts.  Then 
in  figuring  the  portal  stresses  the  ends  of  ports  may  be  considered 
to  lie  in  this  plane  of  contra-flexure.  The  problem  of  finding  the 
stresses  in  a  portal  strut  is  similar  to  that  of  finding  the  wind 
stresses  in  a  roof  truss  supported  on  and  braced  to  steel  columns 
as  explained  in  Art  17, 

The  force  applied  at  top  of  portal  =  2,250  x  3^  =  7,875  Ibs. 

The  horizontal  reaction  at  foot  of  each  post  =  7,875  x  \  =  3,935 
Ibs. 

Since  the  plane  of  contra-flexure  is  assumed  to  be  midway  be- 
tween the  foot  of  posts  and  connection  of  knee  braces,  the  mo- 
ments at  these  points  will  each  be  equal  to  3,935  Ibs.  x  7.5'  = 
29,500  ft.-lbs.  The  moment  at  foot  of  post  is  resisted  by  the  direct 
thrust  in  post  acting  with  a  lever  arm  equal  to  one-half  the  width 
of  bearing  plates.  The  moment  at  knee  brace  connection  is  re- 
sisted by  a  force  at  top  of  post  acting  with  a  lever  arm  of  10  ft. 
Therefore,  this  force  =  29,500  ft.-lbs.  -=-  10'  =  2,950  Ibs.  This 
force  of  2,950  Ibs.  induces  tension  of  the  same  amount  on  leeward 
side  of  top  strut,  and  compression  on  the  windward  side,  but  in  the 
latter  case  the  applied  force  of  7,875  Ibs.  should  be  added.  There- 
fore the  total  compression  on  windward  side  of  top  strut  =  2,950  + 
7,875  =  10,825  Ibs.  The  horizontal  force  at  the  lower  end  of  knee 
braces  is  equal  to  the  induced  force  at  top  of  posts,  plus  the  hori- 
zontal reaction  at  foot  of  posts  =  2,950  +  3,935  =  6,885  Ibs.;  and 
the  stress  in  the  knee  braces  is  equal  to  this  latter  force,  multiplied 
by  length  of  knee  brace  and  divided  by  one-half  the  width  of  portal 
13.2 

=  6,885  x =  10,540  Ibs.    This  stress  will  be  tension  on  the 

8.62 

windward  side  of  portal  and  compression  on  the  leeward  side. 
There  are  no  stresses  in  the  members  shown  in  dotted  lines ;  they 
help  to  stiffen  the  main  members,  and  give  a  more  pleasing  appear- 
ance to  portal.  As  the  stresses  are  all  light,  it  is  only  necessary  to 
proportion  the  members  so  that  their  length  shall  not  exceed  120 
times  their  least  radius  of  gyration. 

The  posts  should  be  proportioned  so  that  the  maximum  fibre 


PIN-CONNECTED  HIGHWAY  SPAN*.  75 

stress  resulting  from  the  combined  action  of  the  dead  load,  live, 
load  and  wind  force  shall  not  exceed  by  more  than  25%  the  permis- 
sible unit  strees  for  dead  and  live  loads  only. 

The  direct  stress  in  post  as  shown  in  Fig.  60  =  56,900  Ibs. 

The  bending  moment  as  above  =  29,500  x  12  =  354,000  inch-lbs. 

The  area  of  the  section  assumed  =  14.31  square  inches,  and  its 
moment  of  resistance  about  an  axis  perpendicular  to  the  cover  plate 

=53-8. 

Then:  Direct  stress  -*-  area  =    56,900  Ibs.  -*-  14.31  =    3,970 
Bending-  moment  -*-  R  =  354,000  in.-lbs.  •*•  53.8  =    6,580 

Total, 10,550  Ibs. 

per  square  inch. 

The  permissible  unit  stress  =  7,870  +  25%  =  9,835  Ibs.  per  sq. 
inch.  Thus  the  post  is  stressed  slightly  too  much  to  fulfill  the  con- 
ditions, but  the  assumed  wind  force  is  probably  much  greater  than 
the  actual. 

The  intermediate  top  struts,  Fig.  65,  are  in  this  case  made  similar 
to  the  portal  struts.  They  are  not  proportioned  for  any  definite 
stress,  but  nevertheless  they  contribute  to  the  stiffness  of  the  bridge, 
and  relieve  the  portal  struts  and  end  posts  from  a  portion  of  their 
wind  stresses. 

Floor  Beam.     Span,  17.25  ft. 
Dead  load  (floor)  =  250  x  15'  =    3,750 
(beam)  650 

4,400 

Live  load  =  1,600  X  15  =  24,000 

Total  (distributed  over  a  length  of  16  ft.)  =  28,400  Ibs. 

Reaction  =  28,400  x  J  =  14,200  Ibs. 

Taking  moments  about  the  center,  M  =  14,200  x  8,625  —  14,200 
X  4  =  65,675  ft.-lbs.,  65,675  X  12  =  788,100  inch-lbs.,  788,100  -*• 
1 5,000==  52.7  ==R. 

A  15"  I  @  42  Ibs.  will  be  used.    R  =  58.9. 

Pin  Moments.  The  pins  in  the  bottom  chord  and  the  one  at  the 
hip  usually  receive  their  maximum  moment  when  the  bridge  is  fully 


76  BRIDGE  AND  STRUCTURAL  DESIGN. 

Lvng     of          Tfu%« 


HO^MOHTM.  5tci\<m  ?m  o     Fig.  66 


S£.CT\OK  ?\r\  B      Fig.  6? 


nORtZOHTM-StCTlOn  Pm   c       Fig.  69 


PIN-CONNECTED  HIGPIWAY  SPAN.  77 

loaded,  but  the  other  pins  in  top  chord  will  be  subjected  to  their 
maximum  moment,  simultaneously  with  the  maximum  stress  in 
main  diagonals  connecting  thereto.  Before  figuring  the  pin  mo- 
ments, it  is  necessary  first  to  find  the  stresses  in  all  the  members 
which  are  connected  by  pin  for  the  condition  of  loading  giving  the 
maximum  moment  on  pin.  The  stresses  in  the  diagonal  members 
should  be  resolved  into  their  vertical  and  horizontal  components, 
and  the  vertical  and  horizontal  moments  figured  separately.  The 
resultant  moment  at  any  point  will  be  equal  to  the  hypothenuse  of 
a  right  angled  triangle  whose  vertical  and  horizontal  sides  are 
equal  respectively  to  the  vertical  and  horizontal  moment  on  pin. 
Having  found  the  maximum  moment,  the  size  of  pin  required  is 
obtained  from  a  table  similar  to  that  in  Carnegie  p.  183,  giving  the 
maximum  bending  moment  allowable  on  pins  of  various  sizes. 
The  values  are  obtained  by  multiplying:  the  R  of  pin  by  the  allow- 
able stress  per  square  inch  on  outer  fibres — in  this  case  22,500  Ibs. 
A  pin  may  be  large  enough  for  the  bending  moment  and  yet  too 
small  for  bearing ;  so  care  must  be  taken  to  ensure  that  the  bearing 
of  all  members  on  the  pin  does  not  exceed  the  allowable  amount, 
which,  in  the  present  example  has  been  taken  at  20,000  Ibs.  per 
square  inch. 

Pin  a,  Fig.  66.     The   vertical   component   of  the  stress   in   end 
post  is  equal  to  this  stress  multiplied  by  depth  of  truss  and  divided 

20 
by  length  of  post  =  56,900  x  —  =  45,500.  The  horizontal  com- 

25 
ponent  is  equal  to  stress  in  post  multiplied  by  length  of  panel  and 

15 
divided  by  length  of  post  =  56,900  x  —  =  34,100  Ibs. 

25 

In  order  to  determine  the  distances  between  the  members,  the 
joint  is  sketched  to  a  large  scale  as  shown.  It  is  only  necessary  to 
consider  the  forces  on  one  side  of  the  centre  line  of  truss,  which 
are  as  follows :  on  the  shoe,  an  upward  force  equal  to  one-half  the 
vertical  component  of  stress  in  end  post  =  22,750  Ibs. ;  on  end 
post,  a  downward  force  of  22,750  Ibs.,  and  a  horizontal  force  acting 
towards  the  left  of  17,050  Ibs. ;  and  on  the  chord  bar,  a  horizontal 
force  to  the  right  of  17,050  Ibs.  The  sum  of  the  forces  acting  in  one 
direction  must  always  equal  the  sum  of  the  forces  acting  in  the 


78  BRIDGE  AND  STRUCTURAL  DESIGN. 

opposite  direction.     The  vertical  and  horizontal  forces  are  shown 
plotted  on  separate  diagrams.    Then: 

Vertical  Momenta.  Horizontal  Moments. 

At  b  =  22,750  X  #"'  =  17,050      At  b  =  17,050  X  %"  =  12,800 

At  c  =  17,050      At  c  =  17,050  X  i#"  =  29,850 

Resultant  Moment. 


At  c  =  -/I7.0503  -f-29,8508  =34.400  inch-lbs. 

Now,  from  table  of  pin  moments  in  column  for  22,500  Ibs.  fibre 
stress,  it  will  be  found  that  a  2^"  pin  has  a  value  of  34,500  inch-lbs. 

As  stated  above,  the  pins  in  the  bottom  chord  and  at  the  hip 
usually  receive  their  maximum  moment  when  the  bridge  is  fully 
loaded.  It  will  therefore  be  necessary  to  figure  the  stresses  in  the 
web  members  for  this  condition  of  loading  before  proceeding 
further.  The  dead  and  live  loads  may  be  taken  together  and  the 
total  stresses  found  in  exactly  the  same  manner  as  for  dead  load 
only.  The  total  panel  load  =  4,000  +  9,000  =  13,000  Ibs.  Reaction 
=  13,000  Ibs.  x  3^  panel  =  45,500  Ibs. 

DEAD  AND  LIVE  LOAD  STRESSES  IN  WEB  MEMBEBS;  BRIDGE  FULLY  LOADED. 


Shear  in  panel  a&=45, 500 —         0=45,500 
"  "     #£=45,500 — 13,000=32,500 

"  «    rc?=45,5oo — 26,000=19,500 

"  "    cfe=45, 500— 39,000=  6,500 


Stress  in  aB  =  45,  SOD  X  f  §  =  56,900 
Be  =  32, 500  X  \\  =  40,600 

"  Cc=*  19,5°°  X  1=19,500 

Cd  =  19, 500  X  f$  =  24,400 

"  Dd=  6,500  X  I  =  6,500 
De=  6,500  XH=  8,100 


For  the  pin  moments  the  stress  in  the  hip  vertical  and  in  all  floor 
beam  hangers  will  be  taken  as  an  ordinary  panel  load,  viz. : 
13,000  Ibs. 

Pin  B.  When  the  bridge  is  fully  loaded,  the  vertical  and  horizon- 
tal components  of  the  stresses  in  the  various  members  connected  by 
this  pin  are  as  follows : 

Members.  Vertical  Components.  Horizontal  Components. 

End  Postal  56,9ooxf£=45»5oo  56,900x^=34,100 

Tie  Bar     Be  40,600x1^=32,500  4o,6ooxif=24'4oo 

Hip  Vertical  Bb  13,000  o 

Top  Chordae  o  58,500 

One-half  of  the  above  vertical  and  horizontal  forces  are  plotted 
in  separate  diagrams,  Fig.  67. 


PIN-CONNECTED  HIGHWAY  SPAN..  79 

Vertical  Moments.  Horizontal  Moments. 

Atb=22,75oxX"  =n,375      At  b=i7,o5oxK  =8,525 

At  0=22,750x1^"  =39,800      At  0=17,050X1^—29,500x1^=6,800 

At  d=22,75ox  3— i6,25oxiX=47,95o      At  d  =6,800 

Resultant  Moment. 
At  d='l/47,-95o~  + 6,800^=48, 500. 

Consequently,  a  2j"  pin  which  has  a  value  of  52,500  inch-lbs.  is 
the  size  required. 

Pin  C,  Fig.  68.  The  condition  for  maximum  moment  on  this  pin 
is  when  the  bridge  is  loaded  from  d  to  h.  The  stresses  C  c  and 
C  d  are  as  shown  on  stress  diagram  Fig.  65.  Since  the  top  chord 
is  continuous  at  this  point,  it  is  only  the  difference  of  the  stresses  in 
B  C  and  C  D  which  need  be  considered,  and  this  is  evidently  equal 
to  the  horizontal  component  of  the  stress  in  C  d. 

The  vertical  and  horizontal  components  of  stresses  in  members 
are  as  follows : 

Members.  Vertical  Components.  Horizontal  Components. 

Top  chord  CO  o  28, 600 X  \\ =17, 150 

Tie  bar  Cd  28, 6ooXfg=22, 900  28,600X^1=17,150 

Post  Cc  22,900                                                o 

Vertical  Moments.  Horizontal  Moments. 

Atb=  o  Atb=8,575Xi^=I0.7oo 

At  c= 1 1, 450X^=10,000  Ate  10,700 

Resultant  Moment. 


At  c=i/io,ooo2  +  io,oo2=i4,6oo  inch-lbs. 


A  2"  pin  which  has  a  bending  value  of  17,700  inch-lbs.  is  the  size 
required.  This  size  may  also  be  used  at  D  and  E. 

Pin  c,  Fig.  69.  When  the  bridge  is  fully  loaded,  the  vertical  and 
horizontal  components  of  the  stresses  in  the  various  members  con- 
nected by  this  pin  are  as  follows  : 

Members.  Vertical  Components.  Horizontal  Components. 

Chord  bar  be  o  34,ioo 

Chord  bar  cc?  o  58,500 

Tie  bar      Be  4o,6ooXf  £=32,500  40,  6ooX  if  =24,400 

Post            Cc  19,500  o 

Floor  beam  hanger  13,000  o 


So 


BRIDGE  AND  STRUCTURAL  DESIGN: 


H-m+H^ 


Centre  Line 


KWiW 


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hanger lo 


y  Chord  jbar  <?/>  jfrtr 


of   Truss      a 


HORIZONTAL  SLCTION  Plrt  ^      Fig-7O 


YERTICAU 
YORCK> 


FORCE* 


I ,.  CenTrg   Line, 


,<&  Try ss 


Tie  bar 


'llCotjinTerp 


IWHNfl 


m° 


bar  cJ      l| 


•—      -  r^tco  q 

lrr±^7|:  '  ''''^  LcViowi  bar  ^~ 
i^^Or^M  2>x% 


r-.  . 

StCTiOK  PVH  d       Y  lg.7l  ,      FORCES 


floorbgam      B  fn^ 


ii 


Centre  Line     «««•»         of    TTUSS     e 


hanger  To 


...      1 


[Chord  bar  ^  3X 


:q_ 


!     Tie  ba.  gf  g^r 


VtRTlCAV 


HIGHWAY  SPAM.  8 1 

Vertical  Moments.  Horizontal  Momenta. 

Atb=                                                         o  At  b=  1 7, 050X1^                           =19,200 

Atc=                                                        o  At  0=17,050X2^— 29,250X1^=  5,6oo 

Atd=i6,25oxi                              16,250  Atd==                                                   5,600 

At  e=i6,25o=4}£— 9,750X3^=36,500  Ate=                                                   5,600 

Resultant  Moment. 
At  e  =  -v/36.5003  +  5,6oo3  =  37.000. 

Thus  a  2§  pin  which  has  a  bending  value  of  40,000  in.-lbs.  is  the 
size  required. 

Pin  bt  Fig.  70.  When  the  bridge  is  fully  loaded,  the  vertical  and 
horizontal  components  of  the  stresses  in  the  various  members  con- 
nected by  this  pin  are  as  follows : 

TVTAmHArfl  Vertical  Horizontal 

Components.  Components. 

Chord  bar  ab                                           o  34,ioo 

Chord  bar  be                                           o  34, 100 

Hip  vertical  Bb  13,000  o 

Floor  beam  hanger  13,000  o 

The  chord  bars  a  b  and  b  c  should  be  spaced  so  that  they  will  pull 
in  as  nearly  a  straight  line  from  a  to  c  as  possible.  By  reference  to 
Fig.  66,  the  distance  from  center  line  of  truss  to  center  line  of  chord 
bar  will  be  found  to  be  3f  " ;  and  in  Fig.  69,  the  center  of  chord  bar 
b  c  is  6f"  from  center  line  of  truss.  The  average  of  these  distances 
is  about  5",  which  should  be  the  distance  to  the  centre  of  the  two 
chord  bars  at  b,  as  shown  in  Fig.  70.  The  object  of  the  spacing  an- 
gle between  the  hip  vertical  and  chord  bar  a  b  is  to  hold  the  top  of 
floor  beam  at  the  proper  distance  below  pin. 

Vertical  Moments.  Horizontal  Moments. 

At  b=                               o  At  b=i7, 050X1^=23,500 

Atc=                                 o  At  c                          =23,500 

At  6=6,500x2^=17. ooo  Atd                        =23,500 

Resultant  Moment. 

At  d  =  •17,000*  -fc  38,500* «-  29,000. 

A  2}^"  pin  which  has  a  bending-  value  of  34,500  inch-lbs.  is  the  size 
required. 

Pin  d,  Fig.  71.    When  bridge  is  fully  loaded,  the  vertical  and 


82  BRIDGE  AND  STRUCTURAL  DESIGN. 

horizontal  components  of  the  stresses  in  the  various  members  con- 
nected by  this  pin  are  as  follows : 

Member.  Vertical  Component.  Horizontal  Component. 

Chord  bar  cd  o  58,50x3 

Chord  bar  de  o  73,ooo 

Tie  bar      Cd  24,4ooxfir=i9>5oo  24, 400 XM^1 4- 500 

Post           Dd  6,500  o 

Counter    dE  o  o 

Floor  beam  hanger  13,000  o 

Vertical  Moments.  Horizontal  Moments. 

Atb=  o  Atb=29,25ox    7/%  =25,600 

Atc=  o  At  0=29,250X1^— 36,500X1=18,500 

Atd=9,75oX%=  8,500  Atd=  18,500 

At  6=9,750x4— 3,250x3^=28,800  Ate=  18,500 

Resultant  Moment. 


At  e  =  1/28, 8oo8  +  i8,5oo3  =  34,200. 

A  2^"  pin  which  has  a  bending  value  of  34,500  inch-lbs.  is  the  size 
required. 

Pin  e,  Fig.  72.  When  the  bridge  is  fully  loaded  the  vertical  and 
horizontal  components  of  the  stresses  in  the  various  members  con- 
nected by  this  pin  are  as  follows : 

Members.  Vertical  Components.  Horizontal  Components. 

Chord  bar   ef  o  73,ooo 

Chord  bar  de  o  73,ooo 

Tie  bar       eF  8,iooxf§  =  6,500  S.iooXsf—  4. 9°° 

Tie  bar      De  8,iooxf5=  6,500  8,iooxif=  4, 900 

Post            Ee  o  o 

Floor  beam  hanger  13,000  o 

Vertical  Moments.  Horizontal  Moments. 

Atb=  o  At  b=36, 500X1—  36,500 

Atc=  o  At  0=36,500X2— 36.500X1=  36,500 

At  £=3,250X1=  3.250  At  d=36, 500X3— 36,500X2+2,450X1=38,900 

At  6=3,250x5+3-250X4=29,250  Ate=  38,900 

Resultant  Moment. 


At  e  =  1/29, 250s  •+-  38,  ooo3  =  48, 700. 

A  2j"  pin  which  has  a  bending  value  of  52,500  inch-lbs.  is  the  size 
required. 

The  size  of  pins  required  at  the  various  panel  points,  as  deter- 


PIN-CONNECTED  HIGHWA  Y  SPAtf.  83 

mined  above,  are  as  follows :  2 J"  at  5  and  e ;  2f "  at  r ;  2j"  at  a,  6 
and  d;  and  2"  at  C,  Z)  and  E.  But,  in  order  to  simplify  the  shop 
work  as  much  as  possible,  3"  pins  will  be  used  throughout,  except 
2£"  pins  at  C,  D  and  E. 

The  pins  in  the  top  chords  and  end  posts  should  be  placed  as  near 
the  centre  of  gravity  of  the  section  as  practicable.  In  the  present 
example  this  centre  of  gravity  will  be  found  to  be  a  little  more  than 
one  inch  above  the  centre  line  of  channels,  but  it  would  be  incon- 
venient to  set  the  pins  at  this  height  as  there  would  not  be  space 
enough  for  some  of  the  bar  heads,  so  they  are  located  one-half  inch 
above  centre  line  of  channels  or  three  inches  below  top  flange.  This 
arrangement  gives  just  room  enough  for  the  largest  bars  at  panel 
point  B,  where  the  space  required  equals  one-half  the  diameter  of 
pin,  plus  the  thickness  of  bars,  =  i%'  +  1^4"  =  27/%" ,  leaving  >6" 
clearance  between  bars  and  cover  plate. 

Shoes,  Rollers  and  Bed  Plates.  Fig.  73..  The  total  load  on  the 
shoes  is  equal  to  the  vertical  component  of  strefss  in  end  post  = 
56,900  X  ||  =  45,500  Ibs.  Then,  since  the  permissible  bearing1 
for  pins  =  20,000  Ibs.  per  square  inch,  45,500  -f-  20,000  =  2.27 
square  inches  required  for  shoe  standards.  Two  7  x  3^  x  -J  Ls  have 
been  used,  and  the  bearing  area  of  pin  on  these  =  -J"  x  3"  x  2  =  3 
square  inches.  This  is  somewhat  more  than  required,  but  it  is  well 
to  make  these  members  quite  stiff.  It  is  difficult  to  determine  the 
exact  thickness  to  make  the  shoe  plates  and  bed  plates,  and  this 
matter  is  usually  left  to  the  judgment  of  the  designer.  In  the 
present  case  these  have  been  made  f"  thick.  The  area  of  the  bed 
plates  should  be  great  enough  so  that  the  pressure  on  the  masonry 
will  not  exceed  300  Ibs.  per  square  inch.  They  must  also  be  large 
enough  to  accommodate  the  rollers  and  the  anchor  bolts.  The 
bearing  area  required  =  45,500-^-300  =  152  square  inches.  The 
area  of  bed  plates  used  =  12"  x  22"  =  264  square  inches. 

The  rollers  should  be  designed  so  that  the  pressure  on  them  per 
lineal  inch  shall  not  exceed  600  V  diameter.  Assuming  2j"  rollers, 
the  permissible  bearing  =  6oc  V  2-5"  =  95°  Ibs.,  then  45,500  -~ 
950  =  48  lineal  inches  required.  There  are  4  rollers  and  the  ef- 
fective length  of  each  =  16"  —  2j"  =  13!".  Then  4  x  13!  =  55 
lineal  inches  provided.  The  rollers  are  turned  down  at  center  to 
pass  over  guide  strips  on  shoe  and  bed  plates,  and  the  ends  are 
turned  down  to  enter  holes  in  the  2\  x  5/16  spacing  bars,  as  shown. 
The  ends  of  the  outer  rollers  are  made  long  enough  to  insert  cotter 


84 


BRIDGE  AND  STRUCTURAL  DESIGN. 


PIN-CONNECTED  HIGHWAY  SPAN,  85 

pins,  but  the  ends  of  the  intermediate  rollers  just  pass  through  the 
spacing  bars. 

The  shoe  plates  are  extended,  as  shown,  to  provide  connections 
for  the  end  laterals,  which  are  made  with  forked  eyes. 

The  fixed  end  bed  plates  are  made  of  cast  iron,  and  their  height 
is  equal  to  the  diameter  of  rollers  plus  the  thickness  of  roller  beds 
-  2^  +  X  =  3#". 

The  anchor  bolts  should  have  sufficient  cross  section  to  resist  the 
total  shear  from  the  assumed  wind  force  in  plane  of  bottom  chords, 
plus  one-half  of  the  wind  force  on  portal  strut.  In  the  table  of 
lateral  stresses  the  dead  and  live  load  shears  in  each  panel  each  = 
2,250  x  3j  =  7,875  Ibs.,  and  the  wind  force  at  top  of  portal  also 
equals  7,875  Ibs.  Then  the  total  force  to  be  resisted  by  anchor 
bolts  =  7,875  X  2^  =  19,700.  The  area  of  two  i#"  bolts  =  2.45 
square  inches,  and  their  shearing  value  =  2.45  x  10,000  Ibs.  = 
24,500  Ibs. 

End  Post  and  Top  Chord.  The  stress  in  end  post  =  56,900  Ibs., 
and  the  permissible  pin  bearing  20,000  Ibs.  per  square  inch,  then 
56,900  -T-  20,000  =  2.84  sq.  inches  required.  The  diameter  of  pins 
=  3",  therefore  thickness  of  bearing  required  =  2.84  -f-  3  =  .95". 
The  thickness  of  the  webs  of  the  7"  [s  @  14.75  Ibs.  =  Vie"-  In  addi- 
tion, Vie"  pin  plates  are  used,  making  the  total  thickness  of  metal 
(yV  +  A)  x  2  =.i^".  The  amount  of  bearing:  on  the  pin  plates 
will  be  in  the  ratio  .of  their  thickness  to  the  total  thickness  of  bear- 

.     f 
ing  =  56,900  x =  23,700.    There  should  be  sufficient  rivets  in 

ii 

pin  plates  to  meet  this  stress.  The  single  shearing  value  of  one  f " 
rivet  =  4,420  Ibs.,  then  23,700  -f-  4,420  =  6  rivets  required ;  8  rivets 
are  shown — 4  in  each  plate.  At  the  upper  end  of  posts,  the  pin  plates 
extend  beyond  the  post,  forming  what  are  called  jaw  plates.  The 
pin  plates  on  top  chord  at  this  point  also  extend  beyond  pin,  but  are 
on  the  inside  of  channels,  as  shown. 

The  stress  in  top  chord  B  C—  58,500  Ibs.,  then  58,500-^20,000 
=  2.92  square  inches  bearing  required;  and  2.92  -f-  3  =  .97"  = 
thickness  of  bearing  required.  The  webs  of  the  7"  [s  at  12.25  Ibs. 
are  5/16"  thick,  and  5/16"  pin  plates  are  used,  making  the  total 
thickness  of  bearing  (5/16  +  Vie)  x  2  —  Ii"-  The  amount  of  bearing 
on  pin  plates  will  then  be  58,500  x  -J  =  29,250  Ibs.  The  number  of 
y  rivets  required  =29,250-1-4420  =  7.  Between  the  end  post 


86  BRIDGE  AND  STRUCTURAL  DESIGN. 

and  top  chord  a  space  of  \n  is  left  to  ensure  that  the  whole  bearing 
will  come  on  pin. 

At  panel  point  C  the  top  chord  is  continuous,  and  it  is  only  nec- 
essary to  provide  sufficient  bearing-  for  the  horizontal  component  of 
the  stress  in  tie  bar  C  d,  which  is  equal  to  28,600  x  Jf  =17,150 
Ibs.  Then  17,150  -=-  20,000  =.85  square  inches  required,  and  .85 
-r-  2.25  =  -.38"  =  thickness  of  bearing  required.  The  webs  of  the 
two  channels  =  f  ",  thus  no  pin  plates  are  required. 

The  top  chord  is  spliced  i'  o"  from  panel  point  D.  Since  this  is 
a  faced  joint,  the  stresses  are  transmitted  directly  by  the  abutting 
surfaces,  and  the  splice  plates  are  required  only  to  hold  the  mem- 
bers in  line. 

The  bent  channels  on  the  top  chord  are  lateral  connections.  The 
lateral  rods,  which  are  J"  diameter,  are  upset  at  the  ends  to  ij", 
and  threaded  for  standard  nuts.  Plate  washers,  f"  thick,  are  used 
to  screw  up  against  the  ends  of  bent  channels. 

Intermediate  Posts.  The  stress  in  post  Cc  =  22,900  Ibs.,  then 
22,900-^-20,000  =  1.14  square  inches  bearing  area  required  on 
pins.  The  diameter  of  the  upper  pin  is  2j",  therefore  1.14-7-2.25 
=  .50  =  thickness  of  bearing  required.  Two  f  "  pin  plates  are  used 
at  top  and  bottom  of  all  posts.  The  posts  must  be  wide  enough  to 
permit  the  floor  beam  hangers  to  pass  between  the  channels  as 
shown.  The  number  of  f"  rivets  required  in  pin  plates  =  22,900 
-5-  3>O7O  =  8,  whereas  there  are  12  rivets  provided. 

Bars.  The  loop  ends  of  the  square  iron  bars  are  made  by  bend- 
ing the  ends  of  the  bar  around  a  pin  and  welding  these  ends  to  the 
body  of  bar.  The  distance  from  centre  of  pin  to  crotch  should  be 
2\  times  the  diameter  of  pin.  The  heads  of  the  flat  steel  bars  are 
made  by  upsetting  the  ends  of  the  bars  and  forging  in  dies  of  the 
required  size.  The  heads  are  made  round  and  of  such  diameter  as 
to  give  an  area  through  centre  line  of  pin  hole  50%  greater  than 
that  of  body  of  bar.  Thus  the  width  of  bars  =  3",  diameter  of  pins 
=  3"  then  3"  +  3"  +  i#"  =  7^"  =  diameter  of  heads. 

Pins  are  turned  down  or  shouldered  at  the  ends  and  provided  with 
chambered  nuts  to  ensure  a  full  bearing  for  the  outer  members. 

Floor  Beams.  The  end  reaction  of  floor  beams  =  16,000  Ibs.  and 
the  unit  stress  for  floor  beam  hangers  =  8,000  Ibs.  per  square  inch, 
therefore  16,000-^-8,000  =  2  square  inches  required  in  hangers. 
The  hangers  are  made  of  i"  square  iron,  bent  to  fit  over  pins  and 
upset  to  i-J"  round  at  ends  and  threaded  for  nuts  as  shown  in  Fig. 


PIN-CONNECTED  HIGHWAY  SPAN. 


88  BRIDGE  AND  STRUCTURAL  DESIGN. 

73.  The  main  nuts  are  of  the  same  thickness  as  diameter  of  thread, 
and  the  lock  nuts  are  i"  thick.  The  top  and  bottom  flanges  of  the 
beams  are  notched  as  shown  in  Fig.  74,  and  on  the  bottom  there  is 
a  washer  plate  4"  x  f  x  8".  The  floor  beams  are  screwed  up  firmly 
against  the  ends  of  the  posts,  which  extend  below  the  pins  at  c,  d 
and  e  far  enough  to  clear  the  largest  bar  heads.  At  panel  point  b, 
where  the  vertical  members  are  rods,  spacing  angles  are  used  to 
hold  the  beam  at  the  same  distance  below  the  pins. 

Bottom  Laterals.  The  end  laterals  which  are  ij"  square  are 
forked  at  one  end  to  connect  with  the  shoe  plate  as  shown  in  Fig.  73. 
At  the  opposite  end  they  are  upset  to  if"  round  and  threaded  for 
standard  nuts..  These  upset  ends  pass  through  holes  in  the  web  of 
floor  beam  b,  and  between  the  lug  angles  which  are  riveted  to  the 
web  of  floor  beam,  and  cut  at  right  angles  to  centre  line  of  lateral 
rods  as  shown  in  Fig.  74.  The  pins  connecting  laterals  with  shoe 
plates  may  be  considered  as  girders  supported  at  both  ends  and 
having  a  concentrated  load  at  the  center  equal  to  the  stress  in 
lateral.  The  distance  centre  to  centre  of  forks  is  about  2",  then 

Wl       21.000X2 

M  = = =  10,500  inch-lbs.,  which  requires  a  if"  pin. 

4  4 

For  uniformity  this  size  will  be  used  throughout. 

The  laterals  in  the  2d  panel,  which  are  i"  square,  have  plain  loop 
eyes  at  one  end  to  connect  with  the  lug  angles  on  floor  beam  b.  At 
the  opposite  end  they  are  upset  to  ij"  round,  and  pass  through 
holes  in  the  web  of  the  floor  beam  c,  and  between  the  lug  angles 
riveted  thereto. 

The  laterals  in  the  3d  panel,  which  are  7/%"  square,  have  plain  loop 
eyes  to  connect  with  the  lug  angles  on  floor  beam  c,  and  are  upset 
to  if"  round  at  opposite  end  and  pass  through  holes  in  web  of 
floor  beam  d,  and  between  the  lug  angles  riveted  thereto. 

In  one  panel  adjacent  to  the  centre  of  the  span  there  will  be  a 
pair  of  laterals  J"  square  with  loop  eyes  at  both  ends.  These  rods 
are  made  in  two  parts  and  provided  with  turnbuckles  for  adjust- 
ment. 

The  lateral  connections  on  the  floor  beams  should  be  placed  as 
near  the  top  flange  and  as  close  to  the  hangers  as  possible. 

The  arrangement  of  the  bottom  laterals  is  shown  in  Fig.  76. 

The  portal  and  intermediate  struts  shown  in  Figs.  74  and  75 
need  no  further  description. 


JO  1 7 


THE  UNIVERSITY  OF  CALIFORNIA  LIBRARY 


